4.1 SOLUTIONS CHAPTER FOUR - De Anza...

4.1 SOLUTIONS 191

CHAPTER FOUR

Solutions for Section 4.1

1. We find a critical point by noting where f ′(x) = 0 or f ′ is undefined. Since the curve is smooth throughout, f ′ is always

defined, so we look for where f ′(x) = 0, or equivalently where the tangent line to the graph is horizontal. These points

are shown in Figure 4.1.

x

f(x)

Figure 4.1

As we can see, there is one critical point. Since it is higher than nearby points, it is a local maximum.

2. We find a critical point by noting where f ′(t) = 0 or f ′ is undefined. Since the curve is smooth throughout, f ′ is always

defined, so we look for where f ′(t) = 0, or equivalently where the tangent line to the graph is horizontal. These points


t

f(t)

A

B

C

D

Figure 4.2

As we can see, there are four labeled critical points. Critical point A is a local maximum because points near it are

all lower; similarly, point B is a local minimum, point C is a local maximum, and point D is a local minimum.



are shown in Figure 4.3:

x

f(x)

Figure 4.3

As we can see, there are three critical points. The leftmost one is a local maximum, because points near it are all

lower; similarly, the middle critical point is surrounded by higher points, and is a local minimum. The critical point to the

right is a local maximum.

192 Chapter Four /SOLUTIONS




x

f(x)

Figure 4.4

As we can see, there is one critical point. Since some nearby points (those to the left) are lower, this point is not a

local minimum; since nearby points to the right are higher, it is not a local maximum. So the one critical point is neither a

local minimum nor a local maximum.

5. (a) One possible answer is shown in Figure 4.5.

(b) One possible answer is shown in Figure 4.6

�local

minima

6

R

localmaximum

Figure 4.5

6

criticalpoints�

Figure 4.6

6. There was a critical point after the first eighteen hours when temperature was at its highest point, a local maximum for the

temperature function.

7. (a)

x

A

B

CE

D f

(b)

x

E

A

B

C

D

g

8. A critical point of f requires f ′(x) = 0 or f ′ undefined. Since f ′ is clearly defined over the relevant range, we find where

f ′(x) = 0, that is, where the graph of f ′ crosses the x-axis. These points are shown and labeled in Figure 4.7.

4.1 SOLUTIONS 193

x

f ′(x)

A B

Figure 4.7

A

B

f(x)

x

Figure 4.8

To the left of critical point A, we see that f ′ > 0 and f is increasing; to the right of the critical point, we see that

f ′ < 0 and f is decreasing. So there is a local maximum at A.

To the left of critical point B, we see that f ′ < 0 and f is decreasing; to the right of the critical point, we see that

f ′ > 0 and f is increasing. So there is a local minimum at B. The sketch of f(x) in Figure 4.8 shows A is a local

maximum and B is a local minimum.

9. A critical point of f would require f ′(x) = 0 or f ′ undefined. Since f ′ is clearly defined over the relevant range, we

wish to find where f ′(x) = 0, or where the graph shown intersects the x-axis. These points are shown and labeled in

Figure 4.9.

x

f ′(x)

A B C

Figure 4.9: The critical points of f(x)

A

B

Cx

f(x)

Figure 4.10: A possible graph of f(x)

To the left of critical point A, f ′ > 0 and f is increasing; to the right, f ′ < 0 and f is decreasing. So there is a local

maximum at A.

To the left of critical point B, f ′ < 0 and f is decreasing; to the right, f ′ > 0 and f is increasing. So there is a local

minimum at B.

To both the left and right of critical point C, f ′ > 0 and so f increases on both sides of point C. So, point C is

neither a local maximum nor a local minimum.

A sketch of f(x) is shown in Figure 4.10.

10. The graph of f in Figure 4.11 appears to be increasing for x < −1.4, decreasing for −1.4 < x < 1.4, and increasing

for x > 1.4. There is a local maximum near x = −1.4 and local minimum near x = 1.4. The derivative of f is

f ′(x) = 3x2 −6. Thus f ′(x) = 0 when x2 = 2, that is x = ±√

2. This explains the critical points near x = ±1.4. Since

f ′(x) changes from positive to negative at x = −√

2, and from negative to positive at x =√

2, there is a local maximum

at x = −√

2 and a local minimum at x =√

2.

−1

1

−6

6

(−1.4, 6.6)

(1.4,−4.6)

x

f(x) = x3 − 6x + 1

Figure 4.11

11. The graph of f in Figure 4.12 appears to be increasing for all x, with no critical points. Since f ′(x) = 3x2 + 6 and

x2 ≥ 0 for all x, we have f ′(x) > 0 for all x. That explains why f is increasing for all x.


−1 1

−8

8

x

f(x) = x3 + 6x + 1

Figure 4.12

12. The graph of f in Figure 4.13 appears to be increasing for x < −1, decreasing for −1 < x < 1 although it is flat at

x = 0, and increasing for x > 1. There are critical points at x = −1 and x = 1, and apparently also at x = 0.

Since f ′(x) = 15x4 − 15x2 = 15x2(x2 − 1), we have f ′(x) = 0 at x = 0,−1, 1. Notice that although f ′(0) = 0,

making x = 0 a critical point, there is no change in sign of f ′(x) at x = 0; the only sign changes are at x = ±1. Thus

the graph of f must alternate increasing/decreasing for x < −1,−1 < x < 1, x > 1, just as we described.

−1 1

−2

2(−1, 2)

(1,−2)

x

f(x) = 3x5 − 5x3

Figure 4.13

13. The graph of f in Figure 4.14 appears to be decreasing for x < 2.3 (almost like a straight line for x < 0), and increasing

sharply for x > 2.3. Here f ′(x) = ex − 10, so f ′(x) = 0 when ex = 10, that is x = ln 10 = 2.302... This is the only

place where f ′(x) changes sign, and it is a minimum of f . Notice that ex is small for x < 0 so f ′(x) ≈ −10 for x < 0,

which means the graph looks like a straight line of slope −10 for x < 0. However, ex gets large quickly for x > 0, so

f ′(x) gets large quickly for x > ln 10, meaning the graph increases sharply there.

(2.3,−13.0)

f(x) = ex − 10x

x

Figure 4.14

14. The graph of f below appears to be decreasing for 0 < x < 0.37, and then increasing for x > 0.37. We have f ′(x) =ln x + x(1/x) = ln x + 1, so f ′(x) = 0 when ln x = −1, that is, x = e−1 ≈ 0.37. This is the only place where f ′

changes sign and f ′(1) = 1 > 0, so the graph must decrease for 0 < x < e−1 and increase for x > e−1. Thus, there is a

local minimum at x = e−1.

(0.37,−0.37)

f(x) = x lnx

x

4.1 SOLUTIONS 195

15. The graph of f in Figure 4.15 looks like a climbing sine curve, alternately increasing and decreasing, with more time

spent increasing than decreasing. Here f ′(x) = 1 + 2 cos x, so f ′(x) = 0 when cos x = −1/2; this occurs when

x = ±2π

3,±4π

3,±8π

3,±10π

3,±14π

3,±16π

3...

Since f ′(x) changes sign at each of these values, the graph of f must alternate increasing/decreasing. However, the

distance between values of x for critical points alternates between (2π)/3 and (4π)/3, with f ′(x) > 0 on the intervals

of length (4π)/3. For example, f ′(x) > 0 on the interval (4π)/3 < x < (8π)/3. As a result, f is increasing on the

intervals of length (4π/3) and decreasing on the intervals of length (2π/3).

(2.1, 3.8)

(4.2, 2.5)6

(8.4, 10.1)

(10.5, 8.7)

x

f(x) = x + 2 sinx

Figure 4.15

16. The function f has critical points at x = 1, x = 3, x = 5.

By the first derivative test, since f ′ is positive to the left of x = 1 and negative to the right, x = 1 is a local maximum.

Since f ′ is negative to the left of x = 3 and positive to the right, x = 3 is a local minimum.

Since f ′ does not change sign at x = 5, this point is neither a local maximum nor a local minimum.

17. The critical points of f occur where f ′ is zero. These two points are indicated in the figure below.

f has crit. pt.Neither max or min

f has alocal min.

f ′(x)

Note that the point labeled as a local minimum of f is not a critical point of f ′.

18. Looking at the graph of the derivative function in Figure 4.16, we see that

f(t) is increasing when 0 < t < 2 or when t > 4, because f ′(t) > 0 for these values of t.f(t) is decreasing when 2 < t < 4, because f ′(t) < 0 for these values of t.f(t) has a local maximum at t = 2, since f(t) is increasing to the left of t = 2 and decreasing to the right of t = 2.

In addition, f(t) has a local maximum at t = 5 since f(t) is increasing to the left of t = 5 and t = 5 is an endpoint.

Also, f(t) has a local minimum at t = 4, since it is decreasing to the left of t = 4 and increasing to the right of t = 4. In

addition, f(t) has a local minimum at t = 0 since it is increasing to the right of t = 0 and t = 0 is an endpoint.

1 2 3 4 5−2

2

6

10

14

t

f ′(t)

Figure 4.16


19. To find the critical points, we set the derivative equal to zero and solve for t.

F ′(t) = Uet + V e−t(−1) = 0

Uet − V

et= 0

Uet =V

et

Ue2t = V

e2t =V

U2t = ln(V/U)

t =ln(V/U)

2.

The derivative F ′(t) is never undefined, so the only critical point is t = 0.5 ln(V/U).

20. (a) The demand for the product is increasing when f ′(t) is positive, and decreasing when f ′(t) is negative. Inspection

of the table suggests that demand is increasing during weeks 0 to 2 and weeks 6 to 10, and decreasing during weeks

3 to 5.

(b) Since f ′(t) = 4 > 0 during week 2 and f ′(t) = −2 < 0 during week 3, the demand for the product changes from

increasing to decreasing near the end of week 2 or the beginning of week 3. Thus the demand has a local maximum

during this time period. Since f ′(t) = −1 < 0 during week 5 and f ′(t) = 3 > 0 during week 6, the demand for the

product changes from decreasing to increasing near the end of week 5 or the beginning of week 6. Thus the demand

has a local minimum during this time period.

21. (a) A critical point occurs when f ′(x) = 0. Since f ′(x) changes sign between x = 2 and x = 3, between x = 6 and

x = 7, and between x = 9 and x = 10, we expect critical points at around x = 2.5, x = 6.5, and x = 9.5.

(b) Since f ′(x) goes from positive to negative at x ≈ 2.5, a local maximum should occur there. Similarly, x ≈ 6.5 is a

local minimum and x ≈ 9.5 a local maximum.

22. Since f ′(x) = 4x3 − 12x2 + 8, we see that f ′(1) = 0, as we expected. We apply the second derivative test to f ′′(x) =12x2 − 24x. Since f ′′(1) = −12 < 0, the graph is concave down at the critical point x = 1, making it a local maximum.

23. Differentiating using the product rule gives

f ′(x) = 3x2(1 − x)4 − 4x3(1 − x)3 = x2(1 − x)3(3(1 − x) − 4x) = x2(1 − x)3(3 − 7x).

The critical points are the solutions to

f ′(x) = x2(1 − x)3(3 − 7x) = 0

x = 0, 1,3

7.

For x < 0, since 1 − x > 0 and 3 − 7x > 0, we have f ′(x) > 0.

For 0 < x < 37

, since 1 − x > 0 and 3 − 7x > 0, we have f ′(x) > 0.

For 37

< x < 1, since 1 − x > 0 and 3 − 7x < 0, we have f ′(x) < 0.

For 1 < x, since 1 − x < 0 and 3 − 7x < 0, we have f ′(x) > 0.

Thus, x = 0 is neither a local maximum nor a local minimum; x = 3/7 is a local maximum; x = 1 is a local

minimum.

24. (a) See Figure 4.17.

?

large a

R

small a

x

Figure 4.17

(b) We see in Figure 4.17 that in each case the graph of f is a parabola with one critical point, its vertex, on the positive

x-axis. The critical point moves to the right along the x-axis as a increases.

4.1 SOLUTIONS 197

(c) To find the critical points, we set the derivative equal to zero and solve for x.

f ′(x) = 2(x − a) = 0

x = a.

The only critical point is at x = a. As we saw in the graph, and as a increases, the critical point moves to the right.


6large a

Ismall a

x

Figure 4.18

(b) We see in Figure 4.18 that in each case f has two critical points placed symmetrically about the origin, one in each

of quadrants II and IV. As a increases, they appear to move farther apart, the one in quadrant II up and to the left, the

one in quadrant IV down and to the right.

(c) To find the critical points, we set the derivative equal to zero and solve for x.

f ′(x) = 3x2 − a = 0

x2 =a

3

x = ±√

a

3.

There are two critical points, at x =√

a/3 and x = −√

a/3. (Since the parameter a is positive, the critical points

exist.) As we saw in the graph, and as a increases, the critical points both move away from the vertical axis.


?

large a?

small a

x

Figure 4.19

(b) We see in Figure 4.19 that in each case f appears to have two critical points. One critical point is a local minimum at

the origin and the other is a local maximum in quadrant I. As the parameter a increases, the critical point in quadrant I

appears to move down and to the left, closer to the origin.

(c) To find the critical points, we set the derivative equal to zero and solve for x. Using the product rule, we have:

f ′(x) = x2 · e−ax(−a) + 2x · e−ax = 0

xe−ax(−ax + 2) = 0

x = 0 and x =2

a.

There are two critical points, at x = 0 and x = 2/a. As we saw in the graph, as a increases the nonzero critical point

moves to the left.


27. Since f(x) has its minimum at x = 3, then x = 3 must be a critical point. So f ′(3) = 0.

f ′(x) = 2x + a so f ′(3) = 6 + a.

Since f ′(3) = 0, then 6 + a = 0 or a = −6. Since (3, 5) is a point on the graph of f(x) we must have f(3) = 5:

f(3) = 32 + a(3) + b = 9 + 3a + b = 5.

Since we know that a = −6, we have:

f(3) = 9 + 3(−6) + b = b − 9 = 5 so b = 14.

Thus, we have found that a = −6, b = 14, giving us f(x) = x2 − 6x + 14.

28. If the minimum of f(x) is at (−2,−3), then the derivative of f must be equal to 0 there. In other words, f ′(−2) = 0. If

f(x) = x2 + ax + b, then

f ′(x) = 2x + a

f ′(−2) = 2(−2) + a = −4 + a = 0

so a = 4. Since (−2,−3) is on the graph of f(x) we know that f(−2) = −3. So

f(−2) = (−2)2 + a(−2) + b = −3

a = 4, so (−2)2 + 4(−2) + b = −3

4 − 8 + b = −3

−4 + b = −3

b = 1

so a = 4 and b = 1, and f(x) = x2 + 4x + 1.

29.

� y = − 12x3

-a = −2

-a = 0

-a = 2

-a = 4

-a = 6

To solve for the critical points, we set dydx

= 0. Since ddx

(

x3 − ax2)

= 3x2 − 2ax, we want 3x2 − 2ax = 0, so

x = 0 or x = 23a. At x = 0, we have y = 0. This first critical point is independent of a and lies on the curve y = − 1

2x3.

At x = 23a, we calculate y = − 4

27a3 = − 1

2

(

23a)3

. Thus the second critical point also lies on the curve y = − 12x3.

30. Recall that the natural logarithm is undefined for x ≤ 0, so the domain of f is x > 0. We see from looking at the graph

of f(x) = x − ln x in the text that this function has one local minimum. We want to assign values to a and b so that this

local minimum occurs at x = 2. The function must therefore have a critical point at x = 2. We find the derivative of

f(x) = a(x − b ln x) and the critical points in terms of a and b.

f ′(x) = a(

1 − b(

1

x

))

= 0

1 − b(

1

x

)

= 0

1 =b

xx = b

We see that f(x) has only one critical point, at x = b. Since we want a critical point at x = 2, we choose b = 2.

4.1 SOLUTIONS 199

Since b = 2, we have f(x) = a(x − 2 lnx). We now use the condition that f(2) = 5 to find a:

f(2) = 5

a(2 − 2 ln 2) = 5

a = 5/(2 − 2 ln 2)

a ≈ 8.147.

We let a = 8.147 and b = 2, so the function is f(x) = 8.147(x − 2 lnx). If we sketch a graph of this function, we see

that this function does indeed have a local minimum approximately at the point (2, 5).

31. We wish to have f ′(3) = 0. Differentiating to find f ′(x) and then solving f ′(3) = 0 for a gives:

f ′(x) = x(aeax) + 1(eax) = eax(ax + 1)

f ′(3) = e3a(3a + 1) = 0

3a + 1 = 0

a = −1

3.

Thus, f(x) = xe−x/3.

32. (a) This function is defined for x > 0. We set the derivative equal to zero and solve for x to find critical points:

f ′(x) = 1 − b1

x= 0

1 =b

xx = b.

The only critical point is at x = b.

(b) Since f ′(x) = 1 − bx−1, the second derivative is

f ′′(x) = bx−2 =b

x2.

Since b > 0, the second derivative is always positive. Thus, the function is concave up everywhere and f has a local

minimum at x = b.

33. (a) The function f(x) is defined for x ≥ 0.

We set the derivative equal to zero and solve for x to find critical points:

f ′(x) = 1 − 1

2ax−1/2 = 0

1 − a

2√

x= 0

2√

x = a

x =a2

4.

Notice that f ′ is undefined at x = 0 so there are two critical points: x = 0 and x = a2/4.

(b) We want the critical point x = a2/4 to occur at x = 5, so we have:

5 =a2

4

20 = a2

a = ±√

20.

Since a is positive, we use the positive square root. The second derivative,

f ′′(x) =1

4ax−3/2 =

1

4

√20x−3/2

is positive for all x > 0, so the function is concave up and x = 5 gives a local minimum. See Figure 4.20.


5

5

x

y

Figure 4.20

34. The function g will have a critical point when g′(x) = 0. Solving this equation gives

g′(x) = 1 − kex = 0

kex = 1

ex =1

k

x = ln1

k= − ln k.

Since the natural logarithm has a domain of all positive real numbers, such a value for x may only exist for k > 0.

35. The domain is all real numbers except x = b. The function is undefined at x = b and has a vertical asymptote there. To

find the critical points, we set the derivative equal to zero and solve for x. Using the quotient rule, we have:

f ′(x) =(x − b)2ax − (ax2)1

(x − b)2= 0

2ax2 − 2abx − ax2

(x − b)2= 0

ax2 − 2abx

(x − b)2= 0.

The first derivative is equal to zero if

ax2 − 2abx = 0

ax(x − 2b) = 0

x = 0 or x = 2b.

There are two critical points: at x = 0 and x = 2b.

36. Since f is differentiable everywhere, f ′ must be zero (not undefined) at any critical points; thus, f ′(3) = 0. Since f has

exactly one critical point, f ′ may change sign only at x = 3. Thus f is always increasing or always decreasing for x < 3and for x > 3. Using the information in parts (a) through (d), we determine whether x = 3 is a local minimum, local

maximum, or neither.

(a) x = 3 is a local maximum because f(x) is increasing when x < 3 and decreasing when x > 3. See Figure 4.21.

3

Figure 4.21

3

Figure 4.22

(b) x = 3 is a local minimum because f(x) heads to infinity to either side of x = 3. See Figure 4.22.

4.2 SOLUTIONS 201

(c) x = 3 is neither a local minimum nor maximum, as f(1) < f(2) < f(4) < f(5). See Figure 4.23.

3

Figure 4.23

3

3

1x

y

Figure 4.24

(d) x = 3 is a local minimum because f(x) is decreasing to the left of x = 3 and must increase to the right of x = 3, as

f(3) = 1 and eventually f(x) must become close to 3. See Figure 4.24.

37. (a) In Figure 4.25, we see that f(θ) = θ − sin θ has a zero at θ = 0. To see if it has any other zeros near the origin, we

use our calculator to zoom in. (See Figure 4.26.) No extra root seems to appear no matter how close to the origin we

zoom. However, zooming can never tell you for sure that there is not a root that you have not found yet.

−2π 2πθ

f(θ) = θ − sin θ

Figure 4.25: Graph of f(θ)

−0.1

0.1θ

f(θ) = θ − sin θ

Figure 4.26: Graph of f(θ)Zoomed In

(b) Using the derivative, f ′(θ) = 1− cos θ, we can argue that there is no other zero. Since cos θ < 1 for 0 < θ ≤ 1, we

know f ′(θ) > 0 for 0 < θ ≤ 1. Thus, f increases for 0 < θ ≤ 1. Consequently, we conclude that the only zero of

f is the one at the origin. If f had another zero at x0, with x0 > 0, then f would have to “turn around”, and recross

the x-axis at x0. But if this were the case, f ′ would be nonpositive somewhere, which we know is not the case.

θ

y

f ′(θ) = 1 − cos θ

Figure 4.27: Graph of f ′(θ)


1. We find an inflection point by noting where the concavity changes. Such points are shown in Figure 4.28. There are three

inflection points.


t

f(t)

Figure 4.28

2. We find an inflection point by noting where the concavity changes. Such points are shown in Figure 4.29. There are two

inflection points.

x

f(x)

Figure 4.29

3. We find an inflection point by noting where the concavity changes. Looking at Figure 4.30, we see that in fact the concavity

changes only at the critical point. So there is one inflection point, which is also a critical point.

f(x)

x

Figure 4.30

4. We find an inflection point by noting where the concavity changes. Such points are shown in Figure 4.31: There are two

inflection points.

x

f(x)

Figure 4.31

5. (a) One possible answer is shown in Figure 4.32.

?

6

)

k

localminima

localmaxima

Figure 4.32

6�

�

inflection points

Figure 4.33

(b) This function is concave down at each local maximum and concave up at each local minimum, so it changes concavity

at least three times. This function has at least 3 inflection points. See Figure 4.33

4.2 SOLUTIONS 203

6. One possible answer is shown in Figure 4.34.

5 10 15 20 25 30

Criticalpoint

Inflectionpoint

Figure 4.34

7. See Figure 4.35.

x

y

Figure 4.35

8. (a) Critical point.

(b) Inflection point.

9. (a) There was a critical point at 6 pm when the temperature was at a local minimum.

(b) The graph of temperature was decreasing but concave up in the morning. In the early afternoon the graph was

decreasing but concave down. There was an inflection point at noon when the northerly wind started blowing. By 6

pm when the temperature was at a local minimum, the graph must have been concave up again so there must have

been a second inflection point between noon and 6 pm. See Figure 4.36.

9 am noon 3 pm 6 pmtime

temperature

Figure 4.36

10. We have f ′(x) = 3x2−36x−10 and f ′′(x) = 6x−36. The inflection point occurs where f ′′(x) = 0, hence 6x−36 = 0.

The inflection point is at x = 6. A graph is shown in Figure 4.37.

6x

f(x)

Figure 4.37


11. A critical point will occur whenever f ′(x) = 0 or f ′ is undefined. Since f ′(x) is always defined, we set

f ′(x) = 2x − 5 = 0

2x = 5

x =5

2

To find the inflection points of f(x), we find where f ′′(x) goes from negative to positive or vice versa. For a point to

satisfy this condition, it must have at least f ′′(x) = 0 or f ′′ undefined. Since f ′′(x) = 2, we know f ′′(x) is always

defined and never equal to zero, so f(x) cannot have inflection points.

So x = 52

is a critical point of f(x), and there are no inflection points.

To identify the nature of the critical point x = 52

that we have found, we can look at a graph of f(x) for values of xnear the critical point. Such a graph is shown in Figure 4.38. From the graph we see that f( 5

2) is a local minimum of f .

2.5 5

(2.5,−3.25)

-local min

x

Figure 4.38


f ′(x) = 3x2 − 3 = 3(x2 − 1) = 0.

Factoring, we get

f ′(x) = 3(x − 1)(x + 1) = 0.

So, x = 1 or x = −1. To find the inflection points of f(x), we find where f ′′(x) goes from negative to positive or vice

versa. For a point to satisfy this condition, it must have at least f ′′(x) = 0 or f ′′ undefined. Since f ′′(x) = 6x, we know

f ′′(x) is always defined. It is zero when 6x = 0, so x = 0. Since f ′′(x) = 6x is negative for x < 0 and positive for

x > 0, x = 0 must be an inflection point for f(x).

So x = 1 and x = −1 are critical points of f(x), and x = 0 is an inflection point for f(x).

To identify the nature of the critical points x = 1 and x = −1 that we have found, we can look at a graph of f(x)for values of x near the critical points. Such a graph is shown in Figure 4.39. From the graph we see that f(−1) is a local

maximum of f and f(1) is a local minimum of f .

(0, 10)� Point of inflection

(1, 8)

-Local max

(−1, 12)f(x)

� Local min

x

y

Figure 4.39

4.2 SOLUTIONS 205

13. f ′(x) = 6x2 + 6x − 36. To find critical points, we set f ′(x) = 0. Then

6(x2 + x − 6) = 6(x + 3)(x − 2) = 0.

Therefore, the critical points of f are x = −3 and x = 2. To find the inflection points of f(x) we look for the points

at which f ′′(x) goes from negative to positive or vice-versa. Since f ′′(x) = 12x + 6, x = −1/2 is an inflection point.

From Figure 4.40, we see the critical point x = −3 is a local maximum and the critical point x = 2 is a local minimum.

-local max

(−3, 86)

� local min(2,−39)

�

point ofinflection

(−.5, 23.5)

x

y

Figure 4.40


f ′(x) =3x2

6+

2x

4− 1 =

x2

2+

x

2− 1 = 0.

Factoring, we get

f ′(x) =1

2(x2 + x − 2) =

1

2(x − 1)(x + 2) = 0.

So x = 1 or x = −2. To find the inflection points of f(x), we find where f ′′(x) goes from negative to positive or vice

versa. For a point to satisfy this condition, it must have at least f ′′(x) = 0 or f ′′ undefined. Since f ′′(x) = x + 12

, we

know f ′′(x) is always defined and is zero when x = − 12

. Since f ′′(x) = x + 12

is negative for x < − 12

and positive for

x > − 12

, x = − 12

must be an inflection point for f(x).

So x = 1 and x = −2 are critical points of f(x), and x = − 12

is an inflection point for f(x).

To identify the nature of the critical points x = 1 and x = −2 that we have found, we can look at a graph of f(x)for values of x near the critical points. Such a graph is shown in Figure 4.41. From the graph we see that f(−2) is a local


(−2, 11/3)� Local max

(

1, 1712

)

� Local min

(

− 12, 6124

)

� Point of inflection

Figure 4.41



f ′(x) = 4x3 − 4x = 4(x3 − x) = 0.

Factoring, we get,

f ′(x) = 4x(x2 − 1) = 4x(x − 1)(x + 1) = 0.

So x = −1, x = 0 or x = 1. To find the inflection points of f(x), we find where f ′′(x) goes from negative to positive or

vice versa. For a point to satisfy this condition, it must have at least f ′′(x) = 0 or f ′′ undefined. Since f ′′(x) = 12x2−4,

we know f ′′(x) is always defined. Factoring, we get

f ′′(x) = 12x2 − 4 = 12(

x2 − 1

3

)

= 12

(

x − 1√3

)(

x +1√3

)

= 0,

which is zero when x = −1/√

3 or x = 1/√

3. Putting in numbers close to −1/√

3, we see that f ′′(x) is positive for

x < −1/√

3 and negative for x > −1/√

3, so x = −1/√

3 must be an inflection point for f(x). Putting in numbers

close to 1/√

3, we see that f ′′(x) is negative for x < 1/√

3 and positive for x > 1/√

3, so x = 1/√

3 must be an

inflection point for f(x).

So x = −1, x = 0 and x = 1 are critical points of f(x), and x = −1/√

3 and x = 1/√

3 are inflection points for

f(x).

To identify the nature of the critical points x = −1, x = 0 and x = 1 that we have found, we can look at a graph of

f(x) for values of x near the critical points. Such a graph is shown in Figure 4.42. From the graph we see that f(−1) is

a local minimum of f , f(0) is a local maximum of f and f(1) is a local minimum of f .

(−1,−1)

-Local min

Local max

(0, 0)

I

Local min

(1,−1)

6

Points ofinflection

I

(

1√3,− 5

9

)(

−1√3,− 5

9

)

f(x)y

Figure 4.42

16. f ′(x) = 12x3 − 12x2. To find critical points, we set f ′(x) = 0. This implies 12x2(x − 1) = 0. So the critical points

of f are x = 0 and x = 1. To find the inflection points of f(x) we look for points at which f ′′(x) goes from negative to

positive or vice-versa. At any such point f ′′(x) is either zero or undefined. Since f ′′(x) = 36x2 − 24x = 12x(3x − 2),

our candidate points are x = 0 and x = 2/3. Both x = 0 and x = 2/3 are inflection points, since at both, f ′′(x) changes

sign.

From Figure 4.43, we see that the critical point x = 1 is a local minimum and the critical point x = 0 is neither a

local maximum nor a local minimum.

4.2 SOLUTIONS 207

−1 0 1 2

1

�

point of inflection

(0, 6)

6

local min

(1, 5)

?

point of inflection

(2/3, 5.4)

x

y

Figure 4.43

17. The derivative of f(x) is f ′(x) = 4x3 − 16x. The critical points of f(x) are points at which f ′(x) = 0. Factoring, we

get

4x3 − 16x = 0

4x(x2 − 4) = 0

4x(x − 2)(x + 2) = 0

So the critical points of f will be x = 0, x = 2, and x = −2.

To find the inflection points of f we look for the points at which f ′′(x) changes sign. At any such point f ′′(x) is

either zero or undefined. Since f ′′(x) = 12x2 − 16 our candidate points are x = ±2/√

3. At both of these points f ′′(x)changes sign, so both of these points are inflection points.

From Figure 4.44, we see that the critical points x = −2 and x = 2 are local minima and the critical point x = 0 is

a local maximum.

−2 2

I

points ofinflection

�

x

-local min(−2,−11)

� local min(2,−11)

� local max(0, 5)

Figure 4.44

18. We first find the critical points of f(x) = x4 − 4x3 + 10. Since f ′(x) = 4x3 − 12x2, setting the derivative equal to 0

and factoring yields

4x3 − 12x2 = 0

4x2(x − 3) = 0

So x = 0 and x = 3 are the critical points of f(x).

We now find the inflection points of f(x). Since

f ′′(x) = 12x2 − 24x,


setting the second derivative equal to 0 and factoring yields

12x2 − 24x = 0

12x(x − 2) = 0.

So x = 0 and x = 2 may be points of inflection of f(x). Since f ′′(x) changes sign at both x = 0 and x = 2, both are

points of inflection for f(x). From Figure 4.45, we see that the critical point x = 3 is a local minimum and the critical

point x = 0 is neither a local minimum nor a local maximum.

(0, 10)

(2,−6)

(3,−17)

� points ofinflection

?

-local min

f(x)

x

y

Figure 4.45


f ′(x) = 5x4 − 20x3 = 5(x4 − 4x3) = 0.

We get

f ′(x) = 5x3(x − 4) = 0,

so x = 0 or x = 4. To find the inflection points of f(x), we find where f ′′(x) goes from negative to positive or vice

versa. For a point to satisfy this condition, it must have at least f ′′(x) = 0 or f ′′ undefined. Since f ′′(x) = 20x3 −60x2,

we know f ′′(x) is always defined. It is zero when 20x3 − 60x2 = 0. Factoring, we get

f ′′(x) = 20(x3 − 3x2) = 20x2(x − 3) = 0.

So x = 0 or x = 3. Putting in values of x very close to zero, we see that f ′′(x) = 20x2(x − 3) is negative for

x < 0 and negative for x > 0, so x = 0 is not an inflection point for f(x). Putting in values close to 3, we see that

f ′′(x) = 20x2(x − 3) is negative for x < 3 and positive for x > 3, so x = 3 is an inflection point for f(x).

So x = 0 and x = 4 are critical points of f(x), and x = 3 is an inflection point for f(x).

To identify the nature of the critical points x = 0 and x = 4 that we have found, we can look at a graph of f(x)for values of x near the critical points. Such a graph is shown in Figure 4.46. From the graph we see that f(0) is a local


4.2 SOLUTIONS 209

�

Local max

(0, 35)

Local min

(4,−221)

?

Point of inflection

(3,−127)

f(x)

x

y

Figure 4.46

20. We see that f ′(x) = 15x4 − 15x2 and f ′′(x) = 60x3 − 30x. Since f ′(x) = 15x2(x2 − 1), the critical points are

x = 0,±1.

To find possible inflection points, we determine when f ′′(x) = 0. Since f ′′(x) = 30x(2x2−1), we have f ′′(x) = 0when x = 0 or x = ±1/

√2. Since f ′′(x) changes sign at each of these points, all are inflection points.

We see from Figure 4.47 that the critical point x = −1 is a local maximum, x = 1 is a local minimum, and x = 0 is

neither.

(−1, 2)?

inflection point

(1,−2)

inflection point

inflection point

6

Figure 4.47

21. From the graph of f(x) in the figure below, we see that the function must have two inflection points. We calculate

f ′(x) = 4x3 + 3x2 − 6x, and f ′′(x) = 12x2 + 6x − 6. Solving f ′′(x) = 0 we find that:

x1 = −1 and x2 =1

2.

Since f ′′(x) > 0 for x < x1, f ′′(x) < 0 for x1 < x < x2, and f ′′(x) > 0 for x2 < x, it follows that both points are

inflection points.

−2

2−2

2

4

Inflection point -

Inflection point�

f(x)

x

22. (a) We set the derivative equal to zero and solve for x to find critical points:

f ′(x) = 4x3 − 4ax = 0

4x(x2 − a) = 0.


We see that there are three critical points:

Critical points: x = 0, x =√

a, x = −√

a.

To find possible inflection points, we set the second derivative equal to zero and solve for x:

f ′′(x) = 12x2 − 4a = 0.

There are two possible inflection points:

Possible inflection points: x =

√

a

3, x = −

√

a

3.

To see if these are inflection points, we determine whether concavity changes by evaluating f ′′ at values on either

side of each of the potential inflection points. We see that

f ′′(−2

√

a

3) = 12(4

a

3) − 4a = 16a − 4a = 12a > 0,

so f is concave up to the left of x = −√

a/3. Also,

f ′′(0) = −4a < 0,

so f is concave down between x = −√

a/3 and x =√

a/3. Finally, we see that

f ′′(2

√

a

3) = 12(4

a

3) − 4a = 16a − 4a = 12a > 0,

so f is concave up to the right of x =√

a/3. Since f(x) changes concavity at x =√

a/3 and x = −√

a/3, both

points are inflection points.

(b) The only positive critical point is at x =√

a, so to have a critical point at x = 2, we substitute:

x =√

a

2 =√

a

a = 4.

Since the critical point is at the point (2, 5), we have

f(2) = 5

24 − 2(4)22 + b = 5

16 − 32 + b = 5

b = 21.

The function is f(x) = x4 − 8x2 + 21.

(c) We have seen that a = 4, so the inflection points are at x =√

4/3 and x = −√

4/3.

23. See Figure 4.48.

x1 x2 x3

x

y

y′′ < 0

y′ > 0

y′′ > 0

y′ > 0

y′′ < 0

y′ > 0

y′′ < 0

y′ < 0

y′′= 0

y′= 0

y′′= 0 y′

= 0

y = f(x)

Figure 4.48

4.2 SOLUTIONS 211


x1 x2 x3

x

yy′′

= 0 y′′= 0 y′′

= 0

y′′ > 0 y′′ < 0 y′′ > 0 y′′ < 0

y = f(x)

y′ < 0 everywhere

Figure 4.49


x1 x2

x

y y′ = 0y′′ = 0 y′′ = 0

y′′ < 0 y′′ > 0 y′′ < 0

y = f(x)

y′ ≥ 0 everywhere

Figure 4.50


x1

y = f(x)

y′′= 0

y′= 2

y′′ > 0

y′ > 0

x

y

Figure 4.51


27. (a)

P (t) =2000

1 + e(5.3−0.4t)

10 20 30 40

500

1000

1500

2000

years since 1774

population of rabbits

From the graph we see that the population levels off at about 2000 rabbits.

(b) The population appears to have been growing fastest when there were about 1000 rabbits, approximately 13 years

after Captain Cook left the original rabbits on the island.

(c) The inflection point coincides with the point of most rapid increase of the rabbit population, that is, the inflection

point occurs approximately 13 years after 1774 when the rabbit population is about 1000 rabbits.

(d) The rabbits reproduce quickly, so their population initially grew very rapidly. Limited food and space availability and

perhaps predators on the island probably account for the population being unable to grow past 2000.

28. (a) Since the volume of water in the container is proportional to its depth, and the volume is increasing at a constant rate,

d(t) = Depth at time t = Kt,

where K is some positive constant. So the graph is linear, as shown in Figure 4.52. Since initially no water is in the

container, we have d(0) = 0, and the graph starts from the origin.

d(t)

t

depth of water

slope = K

Figure 4.52

depth of water

t

d(t)

Figure 4.53

(b) As time increases, the additional volume needed to raise the water level by a fixed amount increases. Thus, although

the depth, d(t), of water in the cone at time t, continues to increase, it does so more and more slowly. This means

d′(t) is positive but decreasing, i.e., d(t) is concave down. See Figure 4.53.


time at which water reacheswidest part of urn

depth of water

time

Figure 4.54

4.2 SOLUTIONS 213


Time at which waterreaches corner of vase

time

depth of water

Figure 4.55


Suppose t1 is the time to fill the left side to the top of the middle ridge. Since the container gets wider as you go up,

the rate dH/dt decreases with time. Therefore, for 0 ≤ t ≤ t1, graph is concave down.

At t = t1, water starts to spill over to right side and so depth of left side does not change. It takes as long for the

right side to fill to the ridge as the left side, namely t1. Thus the graph is horizontal for t1 ≤ t ≤ 2t1.

For t ≥ 2t1, water level is above the central ridge. The graph is climbing because the depth is increasing, but at a

slower rate than for t ≤ t1 because the container is wider. The graph is concave down because width is increasing with

depth. Time t3 represents the time when container is full.

t1 2t1 t3t (time)

H (height)

Figure 4.56

32. (a) The concavity changes at t1 and t3, as shown in Figure 4.57.

f(t)y3

tt3t2t1

y2

y1

Figure 4.57


(b) f(t) grows most quickly where the vase is skinniest (at y3) and most slowly where the vase is widest (at y1). The

diameter of the widest part of the vase looks to be about 4 times as large as the diameter at the skinniest part. Since

the area of a cross section is given by πr2, where r is the radius, the ratio between areas of cross sections at these

two places is about 42, so the growth rates are in a ratio of about 1 to 16 (the wide part being 16 times slower).

33. Since the x3 term has coefficient 1, the polynomial is of the form

y = x3 + ax2 + bx + c.

Differentiating givesdy

dx= 3x2 + 2ax + b.

There is a critical point at x = 2, so dy/dx = 0 at x = 2. Thus

dy

dx

∣

∣

∣

∣

x=2

= 3(22) + 2a(2) + b = 12 + 4a + b = 0, so 4a + b = −12.

We take the second derivative to look for the inflection point. We find

d2y

dx2= 6x + 2a,

and for an inflection point at x = 1, we have 6 + 2a = 0, so a = −3. We now use a = −3 and the relationship

4a + b = −12, which gives 4(−3) + b = −12, so b = 0.We now have

y = x3 − 3x2 + c,

and using the point (1,4) gives

1 − 3 + c = 4,

c = 6.

Thus, y = x3 − 3x2 + 6.

34. Differentiating y = bxe−ax gives

dy

dx= be−ax − abxe−ax = be−ax(1 − ax).

Since we have a critical point at x = 3, we know that 1 − 3a = 0, so a = 1/3.

If b > 0, the first derivative goes from positive values to the left of x = 3 to negative values on the right of x = 3,

so we know this critical point is a local maximum. Since the function value at this local maximum is 6, we have

6 = 3be−3/3 =3b

e,

so b = 2e and

y = 2xe1−x/3.

35. (a) An inflection point occurs whenever the concavity of f(x) changes. If the graph shown is that of f(x), then an

inflection point will occur whenever its concavity changes, or equivalently when the tangent line moves from above

the curve to below or vice-versa. Such points are shown in Figure 4.58.

f(x)

x

Figure 4.58

f ′(x)

x

Figure 4.59

(b) To find inflection points of the function f we must find points where f ′′ changes sign. However, because f ′′ is the

derivative of f ′, any point where f ′′ changes sign will be a local maximum or minimum on the graph of f ′. Such

points are shown in Figure 4.59.

4.3 SOLUTIONS 215

(c) The inflection points of f are the points where f ′′ changes sign. If the graph shown is that of f ′′(x), then we are

looking for where the given graph passes from above the x-axis to below, or vice versa. Such points are shown in

Figure 4.60:

f ′′(x)

x

Figure 4.60

36. (a) This is one of many possible graphs.

(b) Since f must have a bump between each pair of zeros, f could have at most four zeros.

(c) f could well have no zeros at all. To see this, consider the graph of the above function shifted vertically downward.

(d) f must have at least two inflection points. Since f has 3 maxima or minima, it has 3 critical points. Consequently f ′

will have 3 corresponding zeros. Between each consecutive pair of these zeroes f ′ must have a local maximum or

minimum. Thus f ′ will have one local maximum and one local minimum, which implies that f ′′ will have two zeros.

These values, where the second derivative is zero, correspond to points of inflection on the graph of f .

(e) The 3 critical points are zeros of f ′, so degree(f ′) ≥ 3. Thus degree(f) ≥ 4.

(f) For example:

f(x) = −(x + 1)(x − 1)(x − 3)(x − 5)

will look something like the graph in part (a). Many other answers are possible.


1. See Figure 4.61.

1 2 3 4 5 6

10

20

30

40

50f(x)

x

Local max

Local min

Local and global max

Local and global min

Local max

Figure 4.61


2. See Figure 4.62.

1 2 3 4 5

2

4

6

8f(x)

x

Local max


Critical point(not max or min)

Local and global max

Figure 4.62

3. (a) (IV)

(b) (I)

(c) (III)

(d) (II)

4. See Figure 4.63.

3x

y

Figure 4.63

5. See Figure 4.64.

3 8x

y

Figure 4.64

6. See Figure 4.65.

4.3 SOLUTIONS 217

3 8x

y

Figure 4.65

7. See Figure 4.66.

x

y

Figure 4.66

8. True. If the maximum is not at an endpoint, then it must be at critical point of f . But x = 0 is the only critical point of

f(x) = x2 and it gives a minimum, not a maximum.

9. See Figure 4.67.

3 8 10x

y

Figure 4.67


3 10x

y

Figure 4.68



3 8 10x

y

Figure 4.69


10x

y

Figure 4.70

13. We want to maximize the height, y, of the grapefruit above the ground, as shown in the figure below. Using the derivative

we can find exactly when the grapefruit is at the highest point. We can think of this in two ways. By common sense, at the

peak of the grapefruit’s flight, the velocity, dy/dt, must be zero. Alternately, we are looking for a global maximum of y,

so we look for critical points where dy/dt = 0. We have

dy

dt= −32t + 50 = 0 and so t =

−50

−32≈ 1.56 sec.

Thus, we have the time at which the height is a maximum; the maximum value of y is then

y ≈ −16(1.56)2 + 50(1.56) + 5 = 44.1 feet.

1 2 3

20

40

60

t

y

14. (a) We see in Figure 4.71 that the maximum occurs between x = 1 and x = 2 and the maximum value of y is about

y = 30.

4.3 SOLUTIONS 219

(b) The maximum value occurs at a critical point, so we find all critical points of y:

dy

dx= 18 − 10x = 0

x = 1.8.

Since y is a quadratic polynomial with negative leading coefficient, this critical point gives a local maximum, which

is also a global maximum. We find the maximum value of y by substituting x = 1.8:

Maximum value of y = 12 + 18(1.8) − 5(1.8)2 = 28.2.

1.8 3

15

28.2

x

y

Figure 4.71

15. Using a computer to graph the function, f(x) = x3 − ex, and its derivative, f ′(x) = 3x2 − ex, we find that the derivative

crosses the x-axis three times in the interval −1 ≤ x ≤ 4 and twice in the interval −3 ≤ x ≤ 2. See Figures 4.72

and 4.73.

−2 2 4

10

5

−5

−10

−15

−20

f(x)

x

Figure 4.72: Function, f

−2 2 4

20

10

−10

f ′(x)

x

Figure 4.73: Derivative, f ′

Through trial and error, we obtain approximations: local maximum at x ≈ 3.73, local minimum at x ≈ 0.91 and

local maximum at x ≈ −0.46. We can use the approximate values at these points, along with a picture as a guide, to find

the global maximum and minimum on any interval.

(a) We find the global minimum and maximum on the interval −1 ≤ x ≤ 4 by examining the critical points above as

well as the endpoints. Since f(−1) = −1.3679, f(−0.46) = −0.7286, f(0.91) = −1.7308, f(3.73) = 10.2160,

f(4) = 9.4018, we see x ≈ 0.91 gives a global minimum on the interval and x ≈ 3.73 gives a global maximum.

(b) We find the global minimum and maximum on the interval −3 ≤ x ≤ 2 by examining the critical points above as

well as the endpoints. Since f(−3) = −27.0498, f(−0.46) = −0.7286, f(0.91) = −1.7308, f(2) = 0.6109, we

see x = −3 gives a global min and x = 2 a global max. (Even though x ≈ −0.46 gives a local maximum, it does

not give the greatest maximum on this interval; even though x ≈ 0.91 gives a local minimum, it is not the smallest

minimum on this interval.)

16. (a) The maximum photosynthesis rate occurs when t ≈ 50 days.

(b) The leaf is always growing, since the photosynthesis rate is always positive. The leaf is growing fastest when the

photosynthesis rate is greatest, that is t ≈ 50 days.


17. (a) We have

T (D) =(

C

2− D

3

)

D2 =CD2

2− D3

3,

anddT

dD= CD − D2 = D(C − D).

Since, by this formula, dT/dD is zero when D = 0 or D = C, negative when D > C, and positive when D < C,

we have (by the first derivative test) that the temperature change is maximized when D = C.

(b) The sensitivity is dT/dD = CD − D2; its derivative is d2T/dD2 = C − 2D, which is zero if D = C/2, negative

if D > C/2, and positive if D < C/2. Thus by the first derivative test the sensitivity is maximized at D = C/2.

18. (a) Differentiating f(x) = x3 − 3x2 produces f ′(x) = 3x2 − 6x. A second differentiation produces f ′′(x) = 6x − 6.

(b) f ′(x) is defined for all x and f ′(x) = 0 when x = 0, 2. Thus 0 and 2 are the critical points of f .

(c) f ′′(x) is defined for all x and f ′′(x) = 0 when x = 1. When x < 1, f ′′(x) < 0 and when x > 1, f ′′(x) > 0. Thus

the concavity of the graph of f changes at x = 1. Hence x = 1 is an inflection point.

(d) f(−1) = −4, f(0) = 0, f(2) = −4, f(3) = 0. So f has a local maximum at x = 0, a local minimum at x = 2,

global maxima at x = 0 and x = 3, and global minima at x = −1 and x = 2.

(e) Plotting the function f(x) for −1 ≤ x ≤ 3 gives the graph shown in Figure 4.74.

−1 1 2 3

−4

−2

f(x) = x3 − 3x2

x

Figure 4.74

19. (a) Differentiating f(x) = 2x3 − 9x2 + 12x + 1 produces f ′(x) = 6x2 − 18x + 12. A second differentiation produces

f ′′(x) = 12x − 18.

(b) f ′(x) is defined for all x and f ′(x) = 0 when x = 1, 2. Thus x = 1, 2 are critical points.

(c) f ′′(x) is defined for all x and f ′′(x) = 0 when x = 32

. Since the concavity of f changes at this point, it is an

inflection point.

(d) f(−0.5) = −7.5, f(3) = 10, f(1) = 6, f(2) = 5. So f has a local maximum at x = 1 and a local minimum at

x = 2, a global maximum at x = 3 and a global minimum at x = −0.5(e) Plotting the function f(x) for −0.5 ≤ x ≤ 3 gives the graph shown in Figure 4.75.

−1 1 2 3

−10

−5

5

10

f(x) = 2x3 − 9x2 + 12x + 1

x

Figure 4.75

20. (a) Differentiating f(x) = x3 − 3x2 − 9x + 15 produces f ′(x) = 3x2 − 6x − 9. A second differentiation produces

f ′′(x) = 6x − 6.

(b) f ′(x) is defined for all x and f ′(x) = 0 when x = −1, 3. Thus x = −1, 3 are critical points.

(c) f ′′(x) is defined for all x and f ′′(x) = 0 when x = 1. Since the concavity of f changes at this point, it is an

inflection point.

4.3 SOLUTIONS 221

(d) f(−5) = −140, f(4) = −5, f(−1) = 2, f(3) = −12. So f has a global maximum at x = −1 and a global

minimum at x = −5, and a local minimum at x = 3(e) Plotting the function f(x) for −5 ≤ x ≤ 4 gives the graph shown in Figure 4.76:

−4

4

−150f(x) = x3 − 3x2 − 9x + 15

−100

−50

50

x

Figure 4.76

21. (a) Differentiating f(x) = x + sin x produces f ′(x) = 1 + cos x. A second differentiation produces f ′′(x) = − sin x.

(b) f ′(x) is defined for all x and f ′(x) = 0 when x = π. Thus π is the critical point of f .

(c) f ′′(x) is defined for all x and f ′′(x) = 0 when x = 0, x = π and x = 2π. Since the concavity of f changes at each

of these points they are all inflection points.

(d) f(0) = 0, f(2π) = 2π, f(π) = π. So f has a global minimum at x = 0 and a global maximum at x = 2π.

(e) Plotting the function f(x) for 0 ≤ x ≤ 2π gives the graph shown in Figure 4.77:

π 2π

1

2

3

4

5

6

f(x) = x + sinx

x

Figure 4.77

22. (a) Differentiating f(x) = e−x sin x produces f ′(x) = −e−x sin x + e−x cos x. A second differentiation produces

f ′′(x) = −2e−x cos x.

(b) f ′(x) is defined for all x and f ′(x) = 0 when x = π/4 and when x = 5π/4. Thus π/4 and 5π/4 are the critical

points of f .

(c) f ′′(x) is defined for all x and f ′′(x) = 0 when x = π/2, 3π/2. Since the concavity of f changes at both of these

points they are both inflection points.

(d) f(0) = 0, f(2π) = 0, f(π/4) = e−π/4 sin π/4 and f(5π/4) = e−5π/4 sin(5π/4) . So f has a global maximum at

x = π/4 and a global minimum at x = 5π/4.

(e) Plotting the function f(x) for 0 ≤ x ≤ 2π gives the graph shown in Figure 4.78.

π 2π

0.2

0.4

f(x) = e−x sin x

x

Figure 4.78


23. This is a parabola opening downward. We find the critical points by setting g′(x) = 0:

g′(x) = 4 − 2x = 0

x = 2.

Since g′(x) > 0 for x < 2 and g′(x) < 0 for x > 2, the critical point at x = 2 is a local maximum.

As x → ±∞, the value of g(x) → −∞. Thus, the local maximum at x = 2 is a global maximum of g(2) =4 · 2 − 22 − 5 = −1. There is no global minimum. See Figure 4.79.

2

−1 g(x) = 4x − x2 − 5

x

Figure 4.79

24. Differentiating gives

f ′(x) = 1 − 1

x2,

so the critical points satisfy

1 − 1

x2= 0

x2 = 1

x = 1 (We want x > 0).

Since f ′ is negative for 0 < x < 1 and f ′ is positive for x > 1, there is a local minimum at x = 1.

Since f(x) → ∞ as x → 0+ and as x → ∞, the local minimum at x = 1 is a global minimum; there is no global

maximum. See Figure 4.80. The the global minimum is f(1) = 2.

1

2 f(x) = x + 1/x

x

Figure 4.80


g′(t) = 1 · e−t − te−t = (1 − t)e−t,

so the critical point is t = 1.

Since g′(t) > 0 for 0 < t < 1 and g′(t) < 0 for t > 1, the critical point is a local maximum.

As t → ∞, the value of g(t) → 0, and as t → 0+, the value of g(t) → 0. Thus, the local maximum at x = 1 is

a global maximum of g(1) = 1e−1 = 1/e. In addition, the value of g(t) is positive for all t > 0; it tends to 0 but never

reaches 0. Thus, there is no global minimum. See Figure 4.81.

1

1/eg(t) = te−t

t

Figure 4.81

4.3 SOLUTIONS 223

26. Differentiating gives

f ′(x) = 1 − 1

x,


1 − 1

x= 0

1

x= 1

x = 1.

Since f ′ is negative for 0 < x < 1 and f ′ is positive for x > 1, there is a local minimum at x = 1.

Since f(x) → ∞ as x → 0+ and as x → ∞, the local minimum at x = 1 is a global minimum; there is no global

maximum. See Figure 4.82. Thus, the global minimum is f(1) = 1.

1

1 f(x) = x − lnx

x

Figure 4.82

27. Differentiating using the quotient rule gives

f ′(t) =1(1 + t2) − t(2t)

(1 + t2)2=

1 − t2

(1 + t2)2.

The critical points are the solutions to

1 − t2

(1 + t2)2= 0

t2 = 1

t = ±1.

Since f ′(t) > 0 for −1 < t < 1 and f ′(t) < 0 otherwise, there is a local minimum at t = −1 and a local maximum at

t = 1.

As t → ±∞, we have f(t) → 0. Thus, the local maximum at t = 1 is a global maximum of f(1) = 1/2, and the

local minimum at t = −1 is a global minimum of f(−1) = −1/2. See Figure 4.83.

−1

1

−1/2

1/2f(t) = t

1+t2

t

Figure 4.83


f ′(t) = 2 sin t cos t · cos t − (sin2 t + 2) sin t = 0

sin t(2 cos2 t − sin2 t − 2) = 0

sin t(2(1 − sin2 t) − sin2 t − 2) = 0

sin t(−3 sin2 t) = −3 sin3 t = 0.


Thus, the critical points are where sin t = 0, so

t = 0,±π,±2π,±3π, . . . .

Since f ′(t) = −3 sin3 t is negative for −π < t < 0, positive for 0 < t < π, negative for π < t < 2π, and so on, we find

that t = 0,±2π, . . . give local minima, while t = ±π,±3π, . . . give local maxima. Evaluating gives

f(0) = f(±2π) = (0 + 2)1 = 2

f(±π) = f(±3π) = (0 + 2)(−1) = −2.

Thus, the global maximum of f(t) is 2, occurring at t = 0,±2π, . . ., and the global minimum of f(t) is −2, occurring at

t = ±π,±3π, . . . . See Figure 4.84.

−3π−2π−π 3π2ππ

−2

2

t

Figure 4.84

29. We find all critical points:

dy

dx= 2ax + b = 0

x = − b

2a.

Since y is a quadratic polynomial, its graph is a parabola which opens up if a > 0 and down if a < 0. The critical value

is a maximum if a < 0 and a minimum if a > 0.

30. To find the value of w that minimizes S, we set dS/dw equal to zero and solve for w. To find dS/dw, we first solve for

S:

S − 5pw = 3qw2 − 6pq

S = 5pw + 3qw2 − 6pq.

We now find the critical points:

dS

dw= 5p + 6qw = 0

w = −5p

6q.

There is one critical point. Since S is a quadratic function of w with a positive leading coefficient, the function has a

minimum at this critical point.

31. One possible graph of g is in Figure 4.85.

(−2, g(−2))

(0, g(0))

(2, g(2))

g(x)

Figure 4.85

(a) From left to right, the graph of g(x) starts “flat”, decreases slowly at first then more rapidly, most rapidly at x = 0.

The graph then continues to decrease but less and less rapidly until flat again at x = 2. The graph should exhibit

symmetry about the point (0, g(0)).

4.3 SOLUTIONS 225

(b) The graph has an inflection point at (0, g(0)) where the slope changes from negative and decreasing to negative and

increasing.

(c) The function has a global maximum at x = −2 and a global minimum at x = 2.

(d) Since the function is decreasing over the interval −2 ≤ x ≤ 2

g(−2) = 5 > g(0) > g(2).

Since the function appears symmetric about (0, g(0)), we have

g(−2) − g(0) = g(0) − g(2).

32. To find the minimum value of E, we solve

dE

dF= 0.25 − 2(1.7)

F 3= 0

F 3 =2(1.7)

0.25

F =

(

2(1.7)

0.25

)1/3

= 2.4 hours.

Since dE/dF is negative for F < 2.4 and dE/dF is positive for F > 2.4, a foraging time of F = 2.4 hours gives a

local minimum. This is the global minimum for F > 0. See Figure 4.86.

5 10

5

F (hours)

E

Figure 4.86

33. Let w and l be the width and length, respectively, of the rectangular area you wish to enclose. Then

w + w + l = 100 feet

l = 100 − 2w

Area = w · l = w(100 − 2w) = 100w − 2w2

w w

l

WALL

To maximize area, we solve A′ = 0 to find critical points. This gives A′ = 100 − 4w = 0, so w = 25, l = 50. So

the area is 25 · 50 = 1250 square feet. This is a local maximum by the second derivative test because A′′ = −4 < 0.

Since the graph of A is a parabola, the local maximum is in fact a global maximum.

34. (a) Suppose the height of the box is h. The box has six sides, four with area xh and two, the top and bottom, with area

x2. Thus,

4xh + 2x2 = A.

So

h =A − 2x2

4x.


Then, the volume, V , is given by

V = x2h = x2

(

A − 2x2

4x

)

=x

4

(

A − 2x2)

=A

4x − 1

2x3.

(b) The graph is shown in Figure 4.87. We are assuming A is a positive constant. Also, we have drawn the whole graph,

but we should only consider V > 0, x > 0 as V and x are lengths.

√

A2

V = A4

x −12x3

x

V

Figure 4.87

(c) To find the maximum, we differentiate, regarding A as a constant:

dV

dx=

A

4− 3

2x2.

So dV/dx = 0 if

A

4− 3

2x2 = 0

x = ±√

A

6.

For a real box, we must use x =√

A/6. Figure 4.87 makes it clear that this value of x gives the maximum. Evaluating

at x =√

A/6, we get

V =A

4

√

A

6− 1

2

(

√

A

6

)3

=A

4

√

A

6− 1

2· A

6

√

A

6=(

A

6

)3/2

.

35. (a) Since the crow makes n(x) trips to a height of x meters each, the total vertical distance traveled upward is

h = x · n(x) = x +27

xmeters.

(b) At the minimum,

dh

dx= 1 − 27

x2= 0

x2 = 27

x = 5.2 meters.

Since dh/dx is negative for x < 5.2 and dh/dx is positive for x > 5.2, there is a local minimum at x = 5.2. This is

the global minimum for x > 0. See Figure 4.88.

5

5.2

10 15

10

20

30

x (meters)

total upward distance/whelk

Figure 4.88

The dropping height actually observed by biologists is about 5.2 meters.

4.3 SOLUTIONS 227

36. Rewriting the expression for I using the properties of logs gives

I = 192(ln S − ln 762) − S + 763.

Differentiating with respect to S givesdI

dS=

192

S− 1.

At a critical point

192

S− 1 = 0

S = 192.

Sinced2I

dS2= −192

S2,

we see that if S = 192, we have d2I/dS2 < 0, so S = 192 is a local maximum. From Figure 4.89, we see that it is a

global maximum. The maximum possible number of infected children is therefore

I = 192 ln(

192

762

)

− 192 + 763 = 306 children.

192S

I

Figure 4.89

37. (a) Suppose the farmer plants x trees per km2. Then, for x ≤ 200, the yield per tree is 400 kg, so

Total yield, y = 400x kg.

For x > 200, the yield per tree is reduced by 1 kg for each tree over 200, so the yield per tree is 400 − (x− 200) kg

= (600 − x) kg. Thus,

Total yield, y = (600 − x)x = 600x − x2 kg.

The graph of total yield is in Figure 4.90. It is a straight line for x ≤ 200 and a parabola for x > 200.

200 400 600

50,000

100,000

x (trees)

y (kg)

Figure 4.90

(b) The maximum occurs where the derivative of the quadratic is zero, or

dy

dx= 600 − 2x = 0

x = 300 trees/km2.


38. (a) Differentiating twice using the product rule yields dy/dx = (1 − bx)ae−bx and d2y/dx2 = −(2 − bx)abe−bx.

Setting dy/dx = 0, we find that there is a critical point where

1 − bx = 0

x =1

b.

At x = 1b

, we have y = abe−1.

This is a local maximum since at this point d2y/dx2 = −abe−1 < 0.

(b) Since we are concerned only with positive values of x we consider x = 0 and limx→∞ y. When x = 0, y = 0,

and limx→∞ y = 0 also. Since abe−1 is positive, ( 1

b, a

be−1) is in fact a global maximum. This implies that a large

population may actually produce fewer offspring than a smaller one.

39. (a) The variables are S and H . At the maximum value of S, using the product rule, we have

dS

dH= ae−bH + aH(−b)e−bH = 0

ae−bH(1 − bH) = 0

bH = 1

H =1

b.

Thus, the maximum value of S occurs when H = 1/b. To find the maximum value of S, we substitute H = 1/b,

giving

S = a1

be−b(1/b) =

a

be−1.

(b) Increasing a increases the maximum value of S. Increasing b decreases the maximum value of S.

40. (a) When t = 0, we have q(0) = 20(e−0 − e−2·0) = 20(1 − 1) = 0. This is as we would expect: initially there is none

of the drug in the bloodstream.

(b) The maximum value of q(t) occurs where

q′(t) = 20(−e−t + 2e−2t) = 0

−e−t = −2e−2t

e−t

e−2t= 2

et = 2

t = ln 2 = 0.69 hours.

When t = 0.69 hours, we have q(0.69) = 5 mg.

(c) To see what happens in the long run, we let t → ∞. Since e−t → 0 and e−2t → 0, we see that q(t) → 0 as t → ∞.

This is as we would expect: in the long run, the drug will be metabolized or excreted and leave the body.

41. (a) To maximize benefit (surviving young), we pick 10, because that’s the highest point of the benefit graph.

(b) To optimize the vertical distance between the curves, we can either do it by inspection or note that the slopes of the

two curves will be the same where the difference is maximized. Either way, one gets approximately 9.

42. (a) At higher speeds, more energy is used so the graph rises to the right. The initial drop is explained by the fact that the

energy it takes a bird to fly at very low speeds is greater than that needed to fly at a slightly higher speed. When it flies

slightly faster, the amount of energy consumed decreases. But when it flies at very high speeds, the bird consumes a

lot more energy (this is analogous to our swimming in a pool).

(b) f(v) measures energy per second; a(v) measures energy per meter. A bird traveling at rate v will in 1 second travel

v meters, and thus will consume v · a(v) joules of energy in that 1 second period. Thus v · a(v) represents the energy

consumption per second, and so f(v) = v · a(v).

(c) Since v ·a(v) = f(v), a(v) = f(v)/v. But this ratio has the same value as the slope of a line passing from the origin

through the point (v, f(v)) on the curve (see figure). Thus a(v) is minimal when the slope of this line is minimal. To

find the value of v minimizing a(v), we solve a′(v) = 0. By the quotient rule,

a′(v) =vf ′(v) − f(v)

v2.

4.3 SOLUTIONS 229

energy

v

f(v)

a(v)

Thus a′(v) = 0 when vf ′(v) = f(v), or when f ′(v) = f(v)/v = a(v). Since a(v) is represented by the slope

of a line through the origin and a point on the curve, a(v) is minimized when this line is tangent to f(v), so that the

slope a(v) equals f ′(v).

(d) The bird should minimize a(v) assuming it wants to go from one particular point to another, i.e. where the distance

is set. Then minimizing a(v) minimizes the total energy used for the flight.

43. (a) We want the maximum value of I . Using the properties of logarithms, we rewrite the expression for I as

I = k(lnS − lnS0) − S + S0 + I0.

Since k and S0 are constant, differentiating with respect to S gives

dI

dS=

k

S− 1.

Thus, the critical point is at S = k. Since dI/dS is positive for S < k and dI/dS is negative for S > k, we see that

S = k is a local maximum.

We only consider positive values of S. Since S = k is the only critical point, it gives the global maximum value

for I , which is

I = k(ln k − lnS0) − k + S0 + I0.

(b) Since both k and S0 are in the expression for the maximum value of I , both the particular disease and how it starts

influence the maximum.

44. The triangle in Figure 4.91 has area, A, given by

A =1

2x · y =

1

2x3e−3x.

If the area has a maximum, it occurs where

dA

dx=

3

2x2e−3x − 3

2x3e−3x = 0

3

2x2 (1 − x) e−3x = 0

x = 0, 1.

The value x = 0 gives the minimum area, A = 0, for x ≥ 0. Since

dA

dx=

3

2x2(1 − x)e−3x,

we see thatdA

dx> 0 for 0 < x < 1 and

dA

dx< 0 for x > 1.

Thus, x = 1 gives the local and global maximum of

A =1

213e−3·1 =

1

2e3.

-� x

6?y

(x, x2e−3x)

x

y

Figure 4.91


45. (a) The maximum and minimum values of p can be found without taking derivatives, since the function 20 sin(2.5πt)has maximum and minimum values of 20 and −20, respectively. Thus, the maximum value of p is 120 mm Hg and

the minimum value is 80 mm Hg.

(b) The time between successive maxima is the period, which is 2π/(2.5π) = 0.8 seconds.

(c) See Figure 4.92.

1 2

80

100

120-� 0.8 sec

Maximum

Minimum

t (sec)

p(mm Hg)

Figure 4.92

46. (a) If we expect the rate to be nonnegative, then we must have 0 ≤ y ≤ a. See Figure 4.93.

a/2 a

Max rate

y (gm)

rate (gm/sec)

Figure 4.93

(b) The maximum value of the rate occurs at y = a/2, as can be seen from Figure 4.93, or by setting

d

dy(rate) = 0

d

dy(rate) =

d

dy(kay − ky2) = ka − 2ky = 0

y =a

2.

From the graph, we see that y = a/2 gives the global maximum.

47. (a) If we expect the rate to be nonnegative, we must have 0 ≤ y ≤ a and 0 ≤ y ≤ b. Since we assume a < b, we restrict

y to 0 ≤ y ≤ a.

In fact, the expression for the rate is nonnegative for y greater than b, but these values of y are not meaningful

for the reaction. See Figure 4.94.

a b

kab

y (gm)

rate (gm/sec)

Figure 4.94

(b) From the graph, we see that the maximum rate occurs when y = 0; that is, at the start of the reaction.

4.4 SOLUTIONS 231


1. The profit function is positive when R(q) > C(q), and negative when C(q) > R(q). It’s positive for 5.5 < q < 12.5,

and negative for 0 < q < 5.5 and 12.5 < q. Profit is maximized when R(q) > C(q) and R′(q) = C′(q) which occurs

at about q = 9.5. See Figure 4.95.

5 10 15

100

200

300

400C(q)

R(q)-�Profit function

is positive

q (thousands)

$ (thousands)

Figure 4.95

2.

profit

q1 q2

q

marginalcost

q1 q2

qq1 q2

q

marginalrevenue

3. (a) Profit is maximized when R(q)−C(q) is as large as possible. This occurs at q = 2500, where profit = 7500−5500 =$2000.

(b) We see that R(q) = 3q and so the price is p = 3, or $3 per unit.

(c) Since C(0) = 3000, the fixed costs are $3000.

4. (a) The total revenue is given by R = pq so R = 400q − 2q2.(b) The marginal revenue, MR, is given by

dR

dq= 400 − 4q.

When q = 10 the marginal revenue is 360 $/unit.

(c) The total revenue when q = 10 is $3800, and for q = 11 it is $4158 so the increase in revenue is $358 compared

with the marginal revenue of 360 $/unit calculated in part (b).

5. (a) The profit earned by the 51st is the revenue earned by the 51st item minus the cost of producing the 51st item. This

can be approximated by

π′(50) = R′(50) − C′(50) = 84 − 75 = $9.

Thus the profit earned from the 51st item will be approximately $9.

(b) The profit earned by the 91st item will be the revenue earned by the 91st item minus the cost of producing the 91st

item. This can be approximated by

π′(90) = R′(90) − C′(90) = 68 − 71 = −$3.

Thus, approximately three dollars are lost in the production of the 91st item.

(c) If R′(78) > C′(78), production of a 79th item would increase profit. If R′(78) < C′(78), production of one less

item would increase profit. Since profit is maximized at q = 78, we must have

C′(78) = R′(78).


6. (a) See Figure 4.96. We find q1 and q2 by checking to see where the slope of the tangent line to C(q) is equal to the

slope of R(q). Because the slopes of C(q) and R(q) represent marginal cost and marginal revenue, respectively, at

q’s where the slopes are equal, the cost of producing an additional unit of q exactly equals the revenue gained from

selling an additional unit. In other words, q1 and q2 are levels of production at which the additional or marginal profit

from producing an additional unit of q is zero.

q1 q2

C(q)

R(q)

q

$

Figure 4.96

(b) We can think of the vertical distance between the cost and revenue curves as representing the firm’s total profits. The

quantity q1 is the production level at which total profits are at a minimum. We see this by noticing that for points

slightly to the left of q1, the slope of C(q) is slightly greater than the slope of R(q). This means that the cost of

producing an additional unit of q is greater than the revenue earned from selling it. So for points to the left of q1,

additional production decreases profits. For points slightly to the right of q1, the slope of C(q) is less than the slope

of R(q). Thus additional production results in an increase in profits. The production level q1 is the level at which the

firm ceases to take a loss on each additional item and begins to make a profit on each additional item. Note that the

total profit is still negative, and remains so until the graphs cross, i.e, where total cost equals total revenue.

Similar reasoning applies for q2, except it is the level of production at which profits are maximized. For points

slightly to the left of q2, the slope of C(q) is less than the slope of R(q). Thus the cost of producing an additional

unit is less than the revenue gained from selling it. So by selling an additional unit, the firm can increase profits. For

points to the right of q2, the slope of C(q) is greater than the slope of R(q). This means that the profit from producing

and selling an additional unit of q will be negative, decreasing total profits. The point q2 is the level of production at

which the firm stops making a profit on each additional item sold and begins to take a loss.

At both points q1 and q2, note that the vertical distance between C(q) and R(q) is at a local maximum. This

represents the fact that q1 and q2 are local profit minimum and local profit maximum points.

7. (a) At q = 5000, MR > MC, so the marginal revenue to produce the next item is greater than the marginal cost. This

means that the company will make money by producing additional units, and production should be increased.

(b) Profit is maximized where MR = MC, and where the profit function is going from increasing (MR > MC) to

decreasing (MR < MC). This occurs at q = 8000.

8. We know that the maximum (or minimum) profit can occur when

Marginal cost = Marginal revenue or MC = MR.

From the table it appears that MC = MR at q ≈ 2500 and q ≈ 4500. To decide which one corresponds to the maximum

profit, look at the marginal profit at these points. Since

Marginal profit = Marginal revenue − Marginal cost

(or Mπ = MR − MC), we compute marginal profit at the different values of q in Table 4.1:

Table 4.1

q 1000 2000 3000 4000 5000 6000

Mπ = MR − MC −22 −4 4 7 −5 −22

4.4 SOLUTIONS 233

From the table, at q ≈ 2500, we see that profit changes from decreasing to increasing, so q ≈ 2500 gives a local

minimum. At q ≈ 4500, profit changes from increasing to decreasing, so q ≈ 4500 is a local maximum. See Figure 4.97.

Therefore, the global maximum occurs at q = 4500 or at the endpoint q = 1000.

q

q

P ′(q)Mπ

π(q)

1000 2000 3000 4000 5000 6000

1000 2000 4000 5000 6000

10

0

−10

−20

−30

Minimum Profit

Maximum Profit

(1000,−22)

(2000,−4)

(3000, 4)

?(4000, 7)

(5000,−5)

(6000,−22)

Figure 4.97

9. The marginal revenue, MR, is given by differentiating the total revenue function, R. We use the chain rule so

MR =dR

dq=

1

1 + 1000q2· d

dq

(

1000q2)

=1

1 + 1000q2· 2000q.

When q = 10,

Marginal revenue =2000 · 10

1 + 1000 · 102= $0.20/item.

When 10 items are produced, each additional item produced gives approximately $0.20 in additional revenue.

10. (a) Profit = π = R − C; profit is maximized when the slopes of the two graphs are equal, at around q = 350. See

Figure 4.98.

q (quantity)100 200 300 400 500

$C

R

Figure 4.98

MC

MR

q

$/unit

200 400

� Max profit

Figure 4.99

(b) The graphs of MR and MC are the derivatives of the graphs of R and C. Both R and C are increasing everywhere, so

MR and MC are everywhere positive. The cost function is concave down and then concave up, so MC is decreasing

and then increasing. The revenue function is linear and then concave down, so MR is constant and then decreasing.

See Figure 4.99.


11. From the given graph, we can sketch the graph of marginal profit = Mπ = MR − MC: see Figure 4.100.

1000 3000q

$/unit

Mπ = MR − MC

Figure 4.100

1000

3000q

π

π

Figure 4.101

And, from the graph of marginal profit, we can make a sketch showing the shape of the shape of the graph of the profit

function, since the marginal profit curve is the graph of the derivative of the profit function. See Figure 4.101. Because we

don’t know the value of the profit when q = 0, the graph in Figure 4.101 may be shifted vertically.

Since Mπ < 0 for q < 1000 and for q > 3000, while Mπ > 0 for 1000 < q < 3000 we see that profits appear

to be at a minimum when q = 1000 and at a maximum for q = 3000. We see from Figure 4.101 that there is a local

maximum at q = 3000. Therefore, the profit has a global maximum either at q = 3,000 or at the endpoint q = 0.

Because the profit function could be shifted vertically downward, notice that the maximum profit could be zero or

negative.

12. The company should increase production if MR > MC, since increasing production then adds more to revenue than to

cost—a net gain for the company.

(a) Since MC(25) = 17.75 and MR(25) = 30, the company should increase production.

(b) Since MC(50) = 39 and MR(50) = 30, the company should decrease production.

(c) Since MC(80) = 114 and MR(80) = 30, the company should decrease production.

13. We have MR = 30 for all quantities q. The profit is a maximum when MC = MR and MC < MR to the left of the

point and MC > MR to the right of the point. There appear to be two points with MC = MR, one between q = 0 and

q = 10 and another between q = 40 and q = 50.At q = 0 we have MC = 34 > MR, and at q = 10 we have MC = 23 < MR. Thus, the point between q = 0

and q = 10 does not give a maximum profit.

At q = 40 we have MC = 26 < MR, and at q = 50 we have MC = 39 > MR. Thus, the point between q = 40and q = 50 gives a maximum profit.

14. The profit is maximized at the point where the difference between revenue and cost is greatest. Thus the profit is maxi-

mized at approximately q = 4000.

15. (a) We know that marginal cost = C′(q) or the slope of the graph of C(q) at q. Similarly, marginal revenue = R′(q) or

the slope of the graph of R(q) at q. From the graph, the slope of R(q) at q = 3000 is greater than the slope of C(q)at q = 3000. Therefore, marginal revenue is greater than marginal cost at q = 3000. Production should be increased.

(b) The graph of C(q) is steeper than the graph of R(q) at q = 5000. Therefore, marginal cost is greater than marginal

revenue at q = 5000. Production should be decreased.

16. Since marginal revenue is larger than marginal cost around q = 2000, as you produce more of the product your revenue

increases faster than your costs, so profit goes up, and maximal profit will occur at a production level above 2000.

17. First find marginal revenue and marginal cost.

MR = R′(q) = 450

MC = C′(q) = 6q

Setting MR = MC yields 6q = 450, so marginal cost is equal to marginal revenue when

q =450

6= 75 units.

Is profit maximized at q = 75? Profit = R(q) − C(q);

R(75) − C(75) = 450(75) − (10,000 + 3(75)2)

= 33,750 − 26,875 = $6875.

4.4 SOLUTIONS 235

Testing q = 74 and q = 76:

R(74) − C(74) = 450(74) − (10,000 + 3(74)2)

= 33,300 − 26,428 = $6872.

R(76) − C(76) = 450(76) − (10,000 + 3(76)2)

= 34,200 − 27,328 = $6872.

Since profit at q = 75 is more than profit at q = 74 and q = 76, we conclude that profit is maximized locally at q = 75.

The only endpoint we need to check is q = 0.

R(0) − C(0) = 450(0) − (10,000 + 3(0)2)

= −$10,000.

This is clearly not a maximum, so we conclude that the profit is maximized globally at q = 75, and the total profit at this

production level is $6,875.

18. We first need to find an expression for R(q), or revenue in terms of quantity sold. We know that R(q) = pq, where p is

the price of one item. Here p = 45 − 0.01q, so we make the substitution

R(q) = (45 − .01q)q = 45q − 0.01q2.

This is the function we want to maximize. Finding the derivative and setting it equal to 0 yields

R′(q) = 0

45 − 0.02q = 0

0.02q = 45 so

q = 2250.

Is this a maximum?

R′(q) > 0 for q < 2250 and

R′(q) < 0 for q > 2250.

So we conclude that R(q) has a local maximum at q = 2250. Testing q = 0, the only endpoint, R(0) = 0, which is

less than R(2250) = $50,625. So we conclude that revenue is maximized at q = 2250. The price of each item at this

production level is

p = 45 − .01(2250) = $22.50

and total revenue is

pq = $22.50(2250) = $50,625,

which agrees with the above answer.

19. (a) The marginal cost at q = 400 is the slope of the tangent line to C(q) at q = 400. Looking at the graph, we can

estimate a slope of about 1. Thus, the marginal cost is about $1.

(b) At q = 500, we can see that slope of the cost function is greater than the slope of the revenue function. Thus, the

marginal cost is greater than the marginal revenue and thus the 500th item will incur a loss. So, the company should

not produce the 500th item.

(c) The quantity which maximizes profit is at the point where marginal cost equals marginal revenue. This occurs when

the slope of R(q) equals C(q), which occurs at approximately q = 400. Thus, the company should produce about

400 items.

20. (a) Since Profit = Revenue − Cost, we can calculate π(q) = R(q) − C(q) for each of the q values given:

q 0 100 200 300 400 500

R(q) 0 500 1000 1500 2000 2500

C(q) 700 900 1000 1100 1300 1900

π(q) −700 −400 0 400 700 600


We see that maximum profit is $700 and it occurs when the production level q is 400. See Figure 4.102.

0 100 200 300 400 500

500

1000

1500

2000

2500

(0, 700)

(100, 900)(200, 1000)

(300, 1100)

(400, 1300)

(500, 1900)

(100, 500)

(300, 1500)

(400, 2000)

(500, 2500)

?

R(q)

� C(q)

q

Figure 4.102

(b) Since revenue is $500 when q = 100, the selling price is $5 per unit.

(c) Since C(0) = $700, the fixed costs are $700.

21. (a) If q = 3000, the demand equation gives p = 70 − 0.02 · 3000 = 10. That is, at a price of $10, 3000 people attend.

At this price,

Revenue = 3000 people · 10 dollars/person = $30,000.

To find total revenue at a price of $20, first find the attendance at this price. Substituting p = 20 into the demand

equation, p = 70 − 0.02q, gives

20 = 70 − 0.02q.

Solving for q, we get

−50 = −0.02q

2500 = q.

That is, at a price of $20, attendance is 2500 people, and

Revenue = 2500 · 20 = $50, 000.

Notice that, although demand is reduced, the revenue is higher at a price of $20 than at $10.

(b) Since Revenue = Price × Quantity = p · q and p = 70 − 0.002q, we have

R(q) = (70 − 0.02q)q

= 70q − 0.02q2.

(c) To maximize revenue, find the critical points of the revenue function R(q) = 70q − 0.02q2:

R′(q) = 70 − 0.02 · 2q

0 = 70 − 0.04q

70 = 0.04q

1750 = q.

The graph of revenue is a parabola opening downward, so an attendance of 1750 gives the maximum revenue.

(d) Using the demand equation, we find the price corresponding to an attendance of 1750:

p = 70 − 0.02 · 1750 = 70 − 35 = 35.

The optimal price for a ticket at the amusement park is $35.

(e) When the optimal price of $35 is charged, the attendance at the park is 1750 people. Thus, the maximum revenue is

R = pq = 35 · 1750 = $61,250. The corresponding profit cannot be determined without knowing the costs.

4.4 SOLUTIONS 237

22. To maximize revenue, we first must find an expression for revenue in terms of price. We know that R(p) = pq, where

p=price and q=quantity sold. We now need to find an expression for q in terms of p. Using the information given, we find

that

q = 4000 +(4.00 − p)

0.25(200)

Simplification of q yields

q = 4000 + 800(4 − p)

= 4000 + 3200 − 800p

= 7200 − 800p

We can now get an expression for revenue in terms of price.

R(p) = qp = (7200 − 800p)p

= 7200p − 800p2

We want to maximize this function in terms of p. First find the critical points by finding the derivative.

R′(p) = 7200 − 1600p

Setting R′(p) = 0 and solving for p yields

7200 − 1600p = 0

1600p = 7200

p = 4.5

Since R′(p) > 0 for p < 4.5 and R′(p) < 0 for p > 4.5, we conclude that revenue has a local maximum at p = 4.5.

Since this is the only critical point, we conclude that it is the global maximum. So revenue is maximized at a price of

$4.50. The quantity sold at this amount is given by

q = 7200 − 800(4.50) = 3600

and the total revenue is

R(4.5) = 7200(4.5) − 800(4.5)2 = $16,200.

23. We first need to find an expression for revenue in terms of price. At a price of $8, 1500 tickets are sold. For each $1 above

$8, 75 fewer tickets are sold. This suggests the following formula for q, the quantity sold for any price p.

q = 1500 − 75(p − 8)

= 1500 − 75p + 600

= 2100 − 75p.

We know that R = pq, so substitution yields

R(p) = p(2100 − 75p) = 2100p − 75p2

To maximize revenue, we find the derivative of R(p) and set it equal to 0.

R′(p) = 2100 − 150p = 0

150p = 2100

so p = 2100150

= 14. Does R(p) have a maximum at p = 14? Using the first derivative test,

R′(p) > 0 if p < 14 and

R′(p) < 0 if p > 14.

So R(p) has a local maximum at p = 14. Since this is the only critical point for p ≥ 0, it must be a global maximum. So

we conclude that revenue is maximized when the price is $14.


24. (a) In dollars, the revenue earned is R = pq = (−5q + 4000)q. Thus

Profit, π = R − C = (−5q + 4000)q − (6q + 5)

= −5q2 + 3994q − 5.

(b) For maximum profit,dπ

dq= −10q + 3994 = 0,

so

q = 399.4.

(c) For q = 399.4, the profit is given in dollars by

π = −5(399.4)2 + 3994(399.4) − 5 = 797,596.80.

25. (a) Cost C = Fixed cost + Total variable cost = 10,000 + 2q dollars.

(b) We express the demand equation in the form

q = b + mp

where q = 10,100 when p = 5 and q = 12,872 when p = 4.5. Thus

Slope = m =12,872 − 10,100

4.5 − 5=

2772

−0.5= −5544.

To find b, we substitute q = 10,100 and p = 5.

10,100 = b − 5544 · 5.

Thus, b = 37,820, so q = 37,820 − 5544p.

(c) To find the profit as a function of q, we solve for p:

p =37,820

5544− 1

5544q

so

p = 6.822 − 0.00018q.

Then we have

Profit, π = Revenue − Cost = pq − C

= (6.822 − 0.00018q)q − (10,000 + 2q)

= −0.00018q2 + 4.822q − 10,000.

(d) At the maximum profit,dπ

dq= −0.00036q + 4.822 = 0,

so q = 13,394.4. Thus the company should produce 13,394 items. At that production level,

π = −0.00018(13,394)2 + 4.822(13,394) − 10,000 = 22,294 dollars.

26. Consider the rectangle of sides x and y shown in Figure 4.103.

x

y

Figure 4.103

The total area is xy = 3000, so y = 3000/x. Suppose the left and right edges and the lower edge have the shrubs

and the top edge has the fencing. The total cost is

C = 45(x + 2y) + 20(x)

= 65x + 90y.

4.4 SOLUTIONS 239

Since y = 3000/x, this reduces to

C(x) = 65x + 90(3000/x) = 65x + 270,000/x.

Therefore, C′(x) = 65 − 270,000/x2 . We set this to 0 to find the critical points:

65 − 270,000

x2= 0

270,000

x2= 65

x2 = 4153.85

x = 64.450 ft

so that

y = 3000/x = 46.548 ft.

Since C(x) → ∞ as x → 0+ and x → ∞, we see that x = 64.450 is a minimum. The minimum total cost is then

C(64.450) ≈ $8378.54.

27. Let x equal the number of chairs ordered in excess of 300, so 0 ≤ x ≤ 100.

Revenue = R = (90 − 0.25x)(300 + x)

= 27, 000 − 75x + 90x − 0.25x2 = 27, 000 + 15x − 0.25x2

At a critical point dR/dx = 0. Since dR/dx = 15− 0.5x, we have x = 30, and the maximum revenue is $27, 225 since

the graph of R is a parabola which opens downward. The minimum is $0 (when no chairs are sold).

28. (a) Since a/q decreases with q, this term represents the ordering cost. Since bq increases with q, this term represents the

storage cost.

(b) At the minimum,dC

dq=

−a

q2+ b = 0

giving

q2 =a

bso q =

√

a

b.

Sinced2C

dq2=

2a

q3> 0 for q > 0,

we know that q =√

a/b gives a local minimum. Since q =√

a/b is the only critical point, this must be the global

minimum.

29. (a) The business must reorder often enough to keep pace with sales. If reordering is done every t months, then,

Quantity sold in t months = Quantity reordered in each batch

rt = q

t =q

rmonths.

(b) The amount spent on each order is a+ bq, which is spent every q/r months. To find the monthly expenditures, divide

by q/r. Thus, on average,

Amount spent on ordering per month =a + bq

q/r=

ra

q+ rb dollars.

(c) The monthly cost of storage is kq/2 dollars, so

C = Ordering costs + Storage costs

C =ra

q+ rb +

kq

2dollars.


(d) The optimal batch size minimizes C, so

dC

dq=

−ra

q2+

k

2= 0

ra

q2=

k

2

q2 =2ra

k

so

q =

√

2ra

kitems per order.

30. (a) Suppose n passengers sign up for the cruise. If n ≤ 100, then the cruise’s revenue is R = 2000n, so the maximum

revenue is

R = 2000 · 100 = 200,000.

If n > 100, then the price is

p = 2000 − 10(n − 100)

and hence the revenue is

R = n(2000 − 10(n − 100)) = 3000n − 10n2.

To find the maximum revenue, we set dR/dn = 0, giving 20n = 3000 or n = 150. Then the revenue is

R = (2000 − 10 · 50) · 150 = 225,000.

Since this is more than the maximum revenue when n ≤ 100, the boat maximizes its revenue with 150 passengers,

each paying $1500.

(b) We approach this problem in a similar way to part (a), except now we are dealing with the profit function π. If

n ≤ 100, we have

π = 2000n − 80,000 − 400n,

so π is maximized with 100 passengers yielding a profit of

π = 1600 · 100 − 80,000 = $80,000.

If n > 100, we have

π = n(2000 − 10(n − 100)) − (80,000 + 400n).

We again set dπ/dn = 0, giving 2600 = 20n, so n = 130. The profit is then $89,000. So the boat maximizes profit

by boarding 130 passengers, each paying $1700. This gives the boat $89,000 in profit.

31. For each month,

Profit = Revenue − Cost

π = pq − wL = pcKαLβ − wL

The variable on the right is L, so at the maximum

dπ

dL= βpcKαLβ−1 − w = 0

Now β − 1 is negative, since 0 < β < 1, so 1 − β is positive and we can write

βpcKα

L1−β= w

giving

L =(

βpcKα

w

)

11−β

Since β − 1 is negative, when L is just above 0, the quantity Lβ−1 is huge and positive, so dπ/dL > 0. When L is large,

Lβ−1 is small, so dπ/dL < 0. Thus the value of L we have found gives a global maximum, since it is the only critical

point.

4.5 SOLUTIONS 241

32. (a) The units of f(L) are tons/month. If f(1000) = 400, then with 1000 hours of labor in a month the company can

produce 400 tons of product per month.

(b) The units of L are hours/month. For the derivative of f with respect to L we have

Units of f ′(L) =Units of f

Units of L=

tons/month

hours/month=

tons

hour.

If f ′(1000) = 2, then increasing labor from 1000 to 1001 hours in a month increases production by about 2 tons per

month.

(c) At a price of p dollars per ton, Q tons of product costs pQ dollars. One hour’s labor earns w dollars, enough to

purchase Q tons where

w = pQ.

Hence

Real wage = Q =w

p.

(d) The company’s monthly cost is the price of the labor it employs.

C = Hourly wage · Hours of labor = wL.

The monthly revenue is

R = Product price · Quantity = pf(L).

Monthly profit is

π(L) = Revenue − Cost = pf(L) − wL.

(e) At maximum profit, the marginal profit π′(L) is zero. We have

π′(L) = pf ′(L) − w = 0.

Hence pf ′(L) = w and

f ′(L) =w

p,

which means that the marginal product of labor equals the real wage.

The marginal product of labor and the real wage are also equal when the company operates at minimum profit.


1. (a) Since the graph is concave down, the average cost gets smaller as q increases. This is because the cost per item gets

smaller as q increases. There is no value of q for which the average cost is minimized since for any q0 larger than

q the average cost at q0 is less than the average cost at q. Graphically, the average cost at q is the slope of the line

going through the origin and through the point (q, C(q)). Figure 4.104 shows how as q gets larger, the average cost

decreases.

(b) The average cost will be minimized at some q for which the line through (0, 0) and (q, c(q)) is tangent to the cost

curve. This point is shown in Figure 4.105.

q (quantity)

C (cost)

Figure 4.104

q (quantity)

C (cost)

Figure 4.105


2. (a) (i) The average cost of quantity q is given by the formula C(q)/q. So average cost at q = 25 is given by C(25)/25.

From the graph, we see that C(25) ≈ 200, so a(q) ≈ 200/25 ≈ $8 per unit. To interpret this graphically, note

that a(q) = C(q)/q = (C(q)− 0)/(q − 0). This is the formula for the slope of a line from the origin to a point

(q, C(q)) on the curve. So a(25) is the slope of a line connecting (0, 0) to (25, C(25)). See Figure 4.106.

10 20 30 40 50

100

200

300

400

500

C(q)

q

$

Figure 4.106

10 20 30 40 50

100

200

300

400

500

C(q)

q

$

Figure 4.107

(ii) The marginal cost is C′(q). This derivative is the slope of the tangent line to C(q) at q = 25. To estimate

this slope, we draw the tangent line, shown in Figure 4.107. From this plot, we see that the points (50, 300)and (0, 100) are approximately on this line, so its slope is approximately (300 − 100)/(50 − 0) = 4. Thus,

C′(25) ≈ $4 per unit.

(b) We know that a(q) is minimized where a(q) = C′(q). Using the graphical interpretations from parts (i) and (ii), we

see that a(q) is minimized where the line passing from (q, C(q)) to the origin is also tangent to the curve. To find

such points, a variety of lines passing through the origin and the curve are shown in Figure 4.108. The line which is

also a tangent touches the curve at q ≈ 30. So q ≈ 30 units minimizes a(q).

10 20 30 40 50

100

200

300

400

500

C(q)

q

$

Figure 4.108

3. (a) The average cost of quantity q is given by the formula C(q)/q. So average cost at q = 10,000 is given by

C(10,000)/10,000. From the graph, we see that C(10,000) ≈ 16,000, so a(q) ≈ 16,00010,000

≈ $1.60 per unit. The

economic interpretation of this is that $1.60 is each unit’s share of the total cost of producing 10,000 units.

(b) To interpret this graphically, note that a(q) = C(q)q

= C(q)−0q−0

. This is exactly the formula for the slope of a

line from the origin to a point (q, C(q)) on the curve. So a(10,000) is the slope of a line connecting (0, 0) to

(10,000, C(10,000)). Such a line is shown below in Figure 4.109.

4.5 SOLUTIONS 243

10,000

10,000

20,000

q

C(q)

$

Figure 4.109

10,000

10,000

20,000

q

C(q)

$

Figure 4.110

(c) We know that a(q) is minimized where a(q) = C′(q). Using the graphical interpretations from parts (b) and (c), this

is equivalent to saying that the tangent line has the same slope as the line connecting the point on the curve to the

origin. Since these two lines share a point, specifically the point (q, C(q)) on the curve, and have the same slope, they

are in fact the same line. So a(q) is minimized where the line passing from (q, C(q)) to the origin is also tangent to

the curve. To find such points, a variety of lines passing through the origin and the curve are shown in Figure 4.110.

From this plot, we see that the line with the desired properties intersects the curve at q ≈ 18,000. So q ≈ 18,000units minimizes a(q).

4. (a) The marginal cost of producing the qth0 item is simply C′(q0), or the slope of the graph at q0. Since the slope of

the cost function is always 12, the marginal cost of producing the 100th item and the marginal cost of producing the

1000th item is $12.

(b) The average cost at q = 100 is given by

a(100) =C(100)

100=

3700

100= $37.

The average cost at q = 1000 is given by

a(1000) =C(1000)

1000=

14,500

1000= $14.50.

5. The cost function is C(q) = 1000 + 20q. The marginal cost function is the derivative C′(q) = 20, so the marginal cost

to produce the 200th unit is $20 per unit. The average cost of producing 200 units is given by

a(200) =C(200)

200=

5000

200= $25/unit

6. The graph of the average cost function is shown in Figure 4.111.

800

average cost

q

C(q)

q

Figure 4.111


7. (a) The line connecting the origin and the graph of C(q) in Figure 4.112 appears to have minimum slope at q = 6.

Therefore we conclude that average cost is minimized at about q = 6.

6 8

224C(q)

q

$

Figure 4.112

(b) The average cost of the first q items is given by

a(q) =C(q)

q=

q3 − 12q2 + 60q

q= q2 − 12q + 60

We want to minimize a(q). Differentiating gives

a′(q) = 2q − 12.

Setting this equal to 0 and solving yields q = 6. Is this our minimum? We have a′(q) < 0 if q < 6 and a′(q) > 0if q > 6, so q = 6 is a local minimum for a(q). From Figure 4.112, we see that q = 6 is the global minimum for

0 ≤ q ≤ 8.

8. (a) Profit equals revenue minus cost. Your monthly revenue is

1200 slippers × $20/slipper = $24000,

your monthly cost equals 1200 slippers × $2/slipper = $2400. Since you are earning a monthly profit of $24000 −$2400 = $21600, you are making money.

(b) Since additional units produced cost about $3 each, which is above the average cost, producing them increases average

cost. Since additional pairs of slippers cost about $3 to produce and can be sold for $20, you can increase your profit

by making and selling them. This is a case where marginal revenue, which is $20 per slipper, is greater than marginal

cost, which is $3 per pair of slippers.

(c) You should recommend increase in production, since that increases profit. The fact that average cost of production

increases is irrelevant to your decision.

9. (a) a(q) = C(q)/q, so C(q) = 0.01q3 − 0.6q2 + 13q.(b) Taking the derivative of C(q) gives an expression for the marginal cost:

C′(q) = MC(q) = 0.03q2 − 1.2q + 13.

To find the smallest MC we take its derivative and find the value of q that makes it zero. So: MC′(q) = 0.06q−1.2 =0 when q = 1.2/0.06 = 20. This value of q must give a minimum because the graph of MC(q) is a parabola

opening upward. Therefore the minimum marginal cost is MC(20) = 1. So the marginal cost is at a minimum when

the additional cost per item is $1.

(c) a′(q) = 0.02q − 0.6Setting a′(q) = 0 and solving for q gives q = 30 as the quantity at which the average is minimized, since the graph

of a is a parabola which opens upward. The minimum average cost is a(30) = 4 dollars per item.

(d) The marginal cost at q = 30 is MC(30) = 0.03(30)2 − 1.2(30) + 13 = 4. This is the same as the average cost at

this quantity. Note that since a(q) = C(q)/q, we have a′(q) = (qC′(q)− C(q))/q2. At a critical point, q0, of a(q),

we have

0 = a′(q0) =q0C

′(q0) − C(q0)

q20

,

so C′(q0) = C(q0)/q0 = a(q0). Therefore C′(30) = a(30) = 4 dollars per item.

Another way to see why the marginal cost at q = 30 must equal the minimum average cost a(30) = 4 is to view

C′(30) as the approximate cost of producing the 30th or 31st good. If C′(30) < a(30), then producing the 31st

good would lower the average cost, i.e. a(31) < a(30). If C′(30) > a(30), then producing the 30th good would

raise the average cost, i.e. a(30) > a(29). Since a(30) is the global minimum, we must have C′(30) = a(30).

4.5 SOLUTIONS 245

10. (a) The marginal cost tells us that additional units produced would cost about $10 each, which is below the average cost,

so producing them would reduce average cost.

(b) It is impossible to determine the effect on profit from the information given. Profit depends on both cost and revenue,

π = R − C, but we have no information on revenue.

11. (a) N = 100 + 20x, graphed in Figure 4.113.

(b) N ′(x) = 20 and its graph is just a horizontal line. This means that rate of increase of the number of bees with acres

of clover is constant — each acre of clover brings 20 more bees.

On the other hand, N(x)/x = 100/x+20 means that the average number of bees per acre of clover approaches

20 as more acres are put under clover. See Figure 4.114. As x increases, 100/x decreases to 0, so N(x)/x approaches

20 (i.e. N(x)/x → 20). Since the total number of bees is 20 per acre plus the original 100, the average number of

bees per acre is 20 plus the 100 shared out over x acres. As x increases, the 100 are shared out over more acres, and

so its contribution to the average becomes less. Thus the average number of bees per acre approaches 20 for large x.

number of bees

50 100

2100

100 x (acres)

N(x) = 100 + 20x

Figure 4.113

N ′(x) = 20

N(x)x

= 100x

+ 20

50 100

20

x (acres)

bees/acre

i

Figure 4.114

12. (a)Table 4.2

iMarginal Cost

(of ith filter)

Average Cost

(for ith filter)

Marginal

Savings

0 $0 $0 $0

1 $5 $5 $64

2 $6 $5.50 $32

3 $7 $6 $16

4 $8 $6.50 $9

5 $9 $7 $3

6 $10 $7.50 $3

7 $11 $8 $0

(b) She should install four filters. Up to the fourth filter, the marginal cost is less than the marginal savings. This tells

us that for each of the first four filters the developer buys, she will save more on the laundromat than she will have

to pay for the filter. From the fifth filter onward, she pays more for each additional filter than she makes from the

laundromat.

(c) If the rack costs $100 she should not buy it. Instead she should let the laundromat protect itself. This is because, if

she buys even one filter she has to spend $105 on the filter but her savings as a result of the purchase only amount to

$64. If she buys two filters she has to spend $111 on the filters but her total savings as a result of the purchase only

amount to $96. If she buys three filters she has to spend $118 on the filters but her total savings as a result of the

purchase only amount to $112. If she buys four filters she has to spend $126 on the filters but her total savings as a

result of the purchase only amount to $121 etc.

(d) If the rack costs $50 she should buy it and install four filters since she covers the price of the rack with the first

purchase of filters i.e., having bought one filter she will have spent $50 + $5 = $55 but she will have saved $64 on

the laundromat business. Her marginal cost and marginal savings do not change after this point and so she still saves

more on laundromat than she spends on filters if she buys the second, third and fourth filter.


13. It is interesting to note that to draw a graph of C′(q) for this problem, you never have to know what C(q) looks like,

although you could draw a graph of C(q) if you wanted to. By the definition of average cost, we know that C(q) = q·a(q).

Using the product rule we get that C′(q) = a(q) + q · a′(q).

We are given a graph of a(q) which is linear, so a(q) = b + mq, where b = a(0) is the y-intercept and m is the

slope. Therefore

C′(q) = a(q) + q · a′(q) = b + mq + q · m= b + 2mq.

In other words, C′(q) is also linear, and it has twice the slope and the same y–intercept as a(q).

a(q)C′(q)

q

14. Since a(q) = C(q)/q, we use the quotient rule to find

a′(q) =qC′(q) − C(q)

q2=

C′(q) − C(q)/q

q=

C′(q) − a(q)

q.

Since marginal cost is C′, if C′(q) < a(q), then C′(q) − a(q) < 0, so a′(q) < 0.

15. Since a(q) = C(q)/q, we use the quotient rule to find

a′(q) =qC′(q) − C(q)

q2=

C′(q) − C(q)/q

q=

C′(q) − a(q)

q.

Since marginal cost is C′, if C′(q) > a(q), then C′(q) − a(q) > 0, so a′(q) > 0.

16. (a) Differentiating C(q) gives

C′(q) =K

aq(1/a)−1, C′′(q) =

K

a

(

1

a− 1)

q(1/a)−2.

If a > 1, then C′′(q) < 0, so C is concave down.

(b) We have

a(q) =C(q)

q=

Kq1/a + F

q

C′(q) =K

aq(1/a)−1

so a(q) = C′(q) means

Kq1/a + F

q=

K

aq(1/a)−1.

Solving,

Kq1/a + F =K

aq1/a

K(

1

a− 1)

q1/a = F

q =

[

Fa

K(1 − a)

]a

.

4.6 SOLUTIONS 247


1. The effect on the quantity demanded is approximately E times the change in price. A price increase causes a decrease in

quantity demanded and a price decrease causes an increase in quantity demanded.

(a) The quantity demanded decreases by about 0.5(3%) = 1.5%.

(b) The quantity demanded increases by about 0.5(3%) = 1.5%.

2. The effect on the quantity demanded is approximately E times the change in price. A price increase cause a decrease in

quantity demanded and a price decrease cause an increase in quantity demanded.

(a) The quantity demanded decreases by about 2(3%) = 6%.

(b) The quantity demanded increases by about 2(3%) = 6%.

3. (a) We have E = |p/q · dq/dp| = |dollars/tons · tons/dollars|. All the units cancel, and so elasticity has no units.

(b) We have E = |p/q · dq/dp| = |yen/liters · liters/yen|. All the units cancel, and so elasticity has no units.

(c) Elasticity has no units. This is why it makes sense to compare elasticities of different products valued in different

ways and measured in different units. Changing units of measurement will not change the value of elasticity.

4. If shoppers spend 1% more time in the store, then they spend about 1.3% more money.

5. Table 4.5 of Section 4.6 gives the elasticity of peaches as 1.49. Since E > 1, the demand is elastic, and a change in price

will cause a larger percentage change in demand. So if the price of peaches goes up, fewer people will buy them: peaches

are not essential, but a luxury item.

6. Table 4.5 of Section 4.6 gives the elasticity of potatoes as 0.27. Here 0 ≤ E ≤ 1, so the demand is inelastic, meaning that

changes in price will not change the demand so much. This is expected for potatoes; as a staple food item, people will

continue to buy them even if the price goes up.

7. Elasticity will be high. As soon as the price of a brand is raised, many people will switch to another brand causing a drop

in sales.

8. Demand for high-definition TV’s will be elastic, since it is not a necessary item. If the prices are too high, people will not

choose to buy them, so price changes will cause relatively large demand changes.

9. Elasticity will be low. Nearly everyone regards telephone service as a necessity. Since there is not another company they

can turn to, they will keep their telephone service even if the company raises the price. The number of sales will change

very little.

10. The elasticity of demand for a product, E, is given by

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

.

We first find dq/dp = −4p. At a price of $5, the quantity demanded is q = 200 − 50 = 150 and dq/dp = −20, so

E =∣

∣

∣

5

150· (−20)

∣

∣

∣=

2

3.

Since E < 1 demand is inelastic.

11. (a) We use

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

.

So we approximate dq/dp at p = 1.00.

dq

dp≈ 2440 − 2765

1.25 − 1.00=

−325

0.25= −1300

so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

1.00

2765· (−1300)

∣

∣

∣

∣

= 0.470

Since E = 0.470 < 1, demand for the candy is inelastic at p = 1.00.

(b) At p = 1.25,dq

dp≈ 1980 − 2440

1.50 − 1.25=

−460

0.25= −1840


so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

1.25

2440· (−1840)

∣

∣

∣

∣

= 0.943

At p = 1.5,dq

dp≈ 1660 − 1980

1.75 − 1.50=

−320

0.25= −1280

so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

1.50

1980· (−1280)

∣

∣

∣

∣

= 0.970

At p = 1.75,dq

dp≈ 1175 − 1660

2.00 − 1.75=

−485

0.25= −1940

so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

1.75

1660· (−1940)

∣

∣

∣

∣

= 2.05

At p = 2.00,dq

dp≈ 800 − 1175

2.25 − 2.00=

−375

0.25= −1500

so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

2.00

1175· (−1500)

∣

∣

∣

∣

= 2.55

At p = 2.25,dq

dp≈ 430 − 800

2.50 − 2.25=

−370

0.25= −1480

so E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

≈∣

∣

∣

∣

2.25

800· (−1480)

∣

∣

∣

∣

= 4.16

Examination of the elasticities for each of the prices suggests that elasticity gets larger as price increases. In other

words, at higher prices, an increase in price will cause a larger drop in demand than the same size price increase at a

lower price level. This can be explained by the fact that people will not pay too much for candy, as it is somewhat of

a “luxury” item.

(c) Elasticity is approximately equal to 1 at p = $1.25 and p = $1.50.

(d)Table 4.3

p($) qRevenue= p · q ($)

1.00 2765 2765

1.25 2440 3050

1.50 1980 2970

1.75 1660 2905

2.00 1175 2350

2.25 800 1800

2.50 430 1075

We can see that revenue is maximized at p = $1.25, with p = $1.50 a close second, which agrees with part (c).

12. The elasticity of demand is given by

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

.

To evaluate dq/dp we solve p = 90 − 10q for q, so q = (90 − p)/10 and hence

dq

dp= − 1

10.

When p = 50, we find q = 4, dq/dp = −1/10 so

E =∣

∣

∣

50

4·(

− 1

10

)∣

∣

∣= 1.25.

Since the elasticity is

E =Percent change in demand

Percent change in price

4.6 SOLUTIONS 249

when the price increases by 2% the percent change in demand is given by

Percent change in demand = E · Percent change in price

= 2 · (1.25) = 2.5.

Therefore, the percentage change in demand is 2.5%. Since dq/dp < 0 this corresponds to a 2.5% fall in demand.

13. (a) If the price of yams is $2/pound, the quantity sold will be

q = 5000 − 10(2)2 = 5000 − 40 = 4960

so 4960 pounds will be sold.

(b) Elasticity of demand is given by

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· d

dp(5000 − 10p2)

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· (−20p)

∣

∣

∣

∣

=20p2

q

Substituting p = 2 and q = 4960 yields

E =20(2)2

4960=

80

4960= 0.016.

Since E < 1 the demand is inelastic, so it would be more accurate to say “People want yams and will buy them no

matter what the price.”

14. (a) At a price of $2/pound, the quantity sold is

q = 5000 − 10(2)2 = 5000 − 40 = 4960

so the total revenue is

R = pq = 2 · 4960 = $9,920

(b) We know that R = pq, and that q = 5000 − 10p2, so we can substitute for q to find R(p)

R(p) = p(5000 − 10p2) = 5000p − 10p3

To find the price that maximizes revenue we take the derivative and set it equal to 0.

R′(p) = 0

5000 − 30p2 = 0

30p2 = 5000

p2 = 166.67

p = ±12.91

We disregard the negative answer, so p = 12.91 is the only critical point. Is it the maximum? We use the first

derivative test.

R′(p) > 0 if p < 12.91 and

R′(p) < 0 if p > 12.91

So R(p) has a local maximum at p = 12.91. We also test the function at p = 0, which is the only endpoint.

R(0) = 5000(0) − 10(0)3 = 0

R(12.91) = 5000(12.91) − 10(12.91)3 = 64,550 − 21,516.85 = $43,033.15

So we conclude that revenue is maximized at price of $12.91/pound.

(c) At a price of $12.91/pound the quantity sold is

q = 5000 − 10(12.91)2 = 5000 − 1666.68 = 3333.32

so the total revenue is

R = pq = (3333.32)(12.91) = $43,033.16

which agrees with part (b).


(d)

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· d

dp(5000 − 10p2)

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· (−20p)

∣

∣

∣

∣

=20p2

q

Substituting p = 12.91 and q = 3333.32 yields

E =20(12.91)2

3333.32=

3333.36

3333.32≈ 1

which agrees with the result that maximum revenue occurs when E = 1.

15. (a) There were good substitutes for slaves in city occupations - including free blacks. There were no good substitutes in

the countryside.

(b) They were from the countryside, where there was no satisfactory substitute for slaves.

16. The revenue is maximized by finding the critical point of the revenue function:

R = pq = p(1000 − 2p2) = 1000p − 2p3.

Differentiate to find the critical points:

dR

dp= 1000 − 6p2 = 0

p2 =1000

6p ≈ 12.91

To maximize revenues, the price of the product should be $12.91.

17. Demand is elastic at all prices. No matter what the price is, you can increase revenue by lowering the price. In the end,

you would lower your prices all the way to zero. This is not a realistic example, but it is mathematically possible. It would

correspond, for instance, to the demand equation q = 1/p2, which gives revenue R = pq = 1/p which is decreasing for

all prices p > 0.

18. Demand is inelastic at all prices. No matter what the price is, you can increase revenue by raising the price, so there is

no actual price for which your revenue is maximized. This is not a realistic example, but it is mathematically possible.

It would correspond, for instance, to the demand equation q = 1/√

p, which gives revenue R = pq =√

p which is

increasing for all prices p > 0.

19. (a) Since q = k/p, we have dq/dp = −k/p2 and p/q = p/(k/p) = p2/k. Therefore

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

=

∣

∣

∣

∣

p2

k· −k

p2

∣

∣

∣

∣

= 1

Elasticity is a constant equal to 1, independent of the price.

(b) Elasticity equal to 1 corresponds to critical points of the revenue function R. Since R = pq = k is constant,

dR/dp = 0, and so all prices are critical points of the revenue function.

20.

E = |p/q · dq/dp| = |pr+1/k · −kr/pr+1| = r

21. (a) We see that d1 = p0. In addition, we have

dp

dq= Slope of demand curve =

Rise

Run=

−d2

q0

so d2 = −q0 · dp/dq. Therefored1

d2=

p0

−q0 · dpdq

= −p0

q0· dq

dp= E

(b) For prices near the maximum possible, we have d1 > d2 and so elasticity E = d1/d2 is greater than 1 for small

quantities. For prices p near 0 (that is, near the q-axis), we have d1 < d2 and so E = d1/d2 is less than 1 for large

quantities. Elasticity equals 1 where d1 = d2 at exactly half the maximum possible quantity. See Figure 4.115.

4.7 SOLUTIONS 251

E = 1

�

�

E<

1

�

�

E>

1

quantity

price

Figure 4.115

22. Since marginal revenue equals dR/dq and R = pq, we have, using the product rule,

dR

dq=

d(pq)

dq= p · 1 +

dp

dq· q = p

(

1 +q

p· dp

dq

)

= p

(

1 − 1

− pq· dq

dp

)

= p(

1 − 1

E

)

.

23. Marginal cost equals C′(q) = k. On the other hand, by the preceding exercise, marginal revenue equals p(1 − 1/E).

Maximum profit will occur when the two are equal

k = p(1 − 1/E)

Thus

k/p = 1 − 1/E

1/E = 1 − k/p

On the other hand,

Profit

Revenue=

Revenue − Cost

Revenue= 1 − Cost

Revenue= 1 − kq

pq= 1 − k

p=

1

E

24. (a) The approximation EC,q ≈ ∆C/C∆q/q

shows that EC,q measures the ratio of the fractional change in the cost of produc-

tion to the fractional change in the quantity produced. Thus, for example, a 1% increase in production will result in

an EC,q% increase in the cost of production.

(b) Since average cost equals C/q and marginal cost equals dC/dq, we have

Marginal cost

Average cost=

dC/dq

C/q=

q

C· dC

dq= EC,q

25. The approximation Ecross ≈ |∆q/q∆p/p

| shows that the cross-price elasticity measures the ratio of the fractional change in

quantity of chicken demanded to the fractional change in the price of beef. Thus, for example, a 1% increase in the price

of beef will stimulate a Ecross% increase in the demand for chicken, presumably because consumers will react to the price

rise in beef by switching to chicken. The cross-price elasticity measures the strength of this switch.

26. The approximation Eincome ≈∣

∣

∣

∣

∆q/q∆I/I

∣

∣

∣

∣

shows that the income elasticity measures the ratio of the fractional change in

quantity of the product demanded to the fractional change in the income of the consumer. Thus, for example, a 1% increase

in income will translate into an Eincome% increase in the quantity purchased. After an increase in income, the consumer

will tend to buy more. The income elasticity measures the strength of this tendency.


1. (a) As t gets very very large, e−0.08t → 0 and the function becomes P ≈ 40/1. Thus, this model implies that when t is

very large, the population is 40 billion.


(b) A graph of P against t is shown in Figure 4.116.

20 40 60 80

10

20

30

40

t

P (in billions)

Figure 4.116

(c) We are asked to find the time t such that P (t) = 20. Solving we get

20 = P (t) =40

1 + 11e−0.08t

1 + 11e−0.08t =40

20= 2

11e−0.08t = 1

e−0.08t =1

11

ln e−0.08t = ln1

11−0.08t ≈ −2.4

t ≈ −2.4

−0.08= 30

Thus 30 years from 1990 (the year 2020) the population of the world should be 20 billion.

We are asked to find the time t such that P (t) = 39.9. Solving we get

39.9 = P (t) =40

1 + 11e−0.08t

1 + 11e−0.08t =40

39.9= 1.00251

11e−0.08t = 0.00251

e−0.08t =0.00251

11

ln e−0.08t = ln0.00251

11−0.08t ≈ −8.39

t ≈ −8.39

−0.08≈ 105

Thus 105 years from 1990 (the year 2095) the population of the world should be 39.9 billion.

2. (a) Figure 4.117 shows the rate of sales as a function of time. The point of diminishing returns is reached when the rate

of sales is at a maximum; this happens during the 4th month.

1 2 3 4 5 60

100

200

300

400

500

600

700

800

month

rate of sales

Figure 4.117

4.7 SOLUTIONS 253

(b) Total sales for the first four months are

140 + 520 + 680 + 750 = 2090 sales.

(c) Assuming logistic growth, the limiting value should be twice the value at the point of diminishing returns, so total

sales should be 2(2090) = 4180 sales.

3. (a) The data is plotted in Figure 4.118. On the intervals less than 10, the average rate of change increases, and greater

than 10, this rate decreases, so the point of diminishing returns is about 10.

2 6 10 14

30

60 Point ofdiminishing

returns

months sincegame introduced

total sales(thousands)

Figure 4.118

(b) The point of diminishing returns happens when total sales reach roughly 30,000. This predicts total sales of about

2(30) = 60,000.

4. (a) Substituting the value t = 0 we get

N(0) =400

1 + 399e−0.4(0)

=400

1 + 399(1)

= 1

The fact that N(0) = 1 tells us that at the moment the rumor begins spreading, there is only one person who knows

the content of the rumor.

(b) Substituting t = 2 we get

N(2) =400

1 + 399e−0.4(2)

≈ 400

1 + 399(0.449)

≈ 2

Substituting in t = 10 we get

N(10) =400

1 + 399e−0.4(10)

=400

1 + 7.308≈ 48

(c) The graph of N(t) is shown in Figure 4.119.

5 10 15 20

200

400

t

N(t)

Figure 4.119


(d) We are asked to find the time t at which 200 people will have heard the rumor. In other words, we are asked to solve

200 =400

1 + 399e−0.4t.

Solving we get

1 + 399e−0.4t =400

200= 2

399e−0.4t = 1

e−0.4t =1

399

ln e−0.4t = ln1

399−0.4t ≈ −5.989

t =−5.989

−0.4≈ 15.

Thus after 15 hours half the people will have heard the rumor.

We are asked to solve for the time at which virtually everyone will have heard the rumor. Since the value of

N(t) only makes sense after rounding to the nearest integer (i.e., it does not make sense to say that 2.5 people heard

the rumor,) asking for the time at which all 400 people will have heard the rumor is equivalent to asking for the time

t such that

N(t) = 399.5.

Solving this we get

399.5 = N(t) =400

1 + 399e−0.4t

1 + 399e−0.4t =400

399.5≈ 1.00125

399e−0.4t ≈ 0.00125

e−0.4t ≈ 0.00125

399

ln e−0.4t ≈ ln0.00125

399−0.4t ≈ −12.7

t ≈ −12.7

−0.4= 31.75

Thus after approximately 32 hours virtually everyone will have heard the rumor.

(e) The rumor is spreading fastest at L/2 = 400/2 = 200 or when 200 people have already heard the rumor, so after

about 15 hours.

5.

−15 10

10

0t

P

Figure 4.120: C = 0.1

−15 10

10

0t

P

Figure 4.121: C = 1

−15 10

10

0t

P

Figure 4.122: C = 10

See Figures 4.120–4.122. The value of C appears to affect where the curve cross the vertical axis. When C = 1, the

graph crosses the vertical axis at t = 0 or at the point of diminishing returns. If C > 1, the graph crosses the the vertical

axis before the point of diminishing returns; the higher C is, the sooner the graph crosses the vertical axis. If C < 1, the

curve crosses the vertical axis after it changes concavity; the smaller C is, the closer to the limiting value the graph is

before crossing the vertical axis.

4.7 SOLUTIONS 255

6. Sales of a new product could very well follow a logistic curve. At first, sales will grow exponentially as more and more

people hear of the product and decide to buy it. Eventually, though, everyone will know about the product and while

people may still buy it, not as many will (sales will slow down). Eventually, it’s possible that everyone who would buy

the product already has, at which point sales will stop. It behooves the seller to notice the point of diminishing returns so

they don’t make more of their product than people will want to buy.

7. (a) The limiting value of f(t) is about 36, so the total number of infected computers is about 36,000.

(b) The curve has an inflection point at about t ≈ 16 hours, and then n = f(16) ≈ 18.

(c) The virus was spreading fastest at about when t = 16, that is at 4 pm on July 19, 2001. At that time, about 18,000computers were infected.

(d) At the inflection point, the number of computers infected is about half the number infected in the long run.

8. The slope of the curve, dy/dt, is given by

dy

dt= −50

(

1 + 6e−2t)−2 (−12e−2t

)

= 600e−2t(

1 + 6e−2t)−2

.

If the slope has a maximum, it occurs where its derivative is zero, that is where

d2y

dt2= 600

(

−2e−2t) (

1 + 6e−2t)−2 − 1200e−2t

(

1 + 6e−2t)−3 (−12e−2t

)

= 0

1200e−2t

(1 + 6e−2t)3

(

−(1 + 6e−2t) + 12e−2t)

= 0

6e−2t − 1 = 0

e−2t =1

6

−2t = ln(

1

6

)

t = −1

2ln(

1

6

)

=1

2ln 6.

For this value of t,

y =50

1 + 6e−2( 12

ln 6)=

50

1 + 6e− ln 6=

50

1 + 6(

16

) = 25.

Since we see in Figure 4.123 that the slope increases as t increases from 0 and tends to 0 as t → ∞, the only critical point

t = 12

ln 6 is a local and global maximum for the slope.

12

ln 6

25

50 y =50

1 + 6e−2t

t

y

Figure 4.123

9. (a), (b) The graph of P (t) with carrying capacity L and point of diminishing returns t0 is in Figure 4.124. The derivative

P ′(t) is also shown.


t0

L

t

P (t)

P ′(t)

P (t0)

Figure 4.124

(c) Keeping track of rate of sales is the same as keeping track of the derivative P ′(t). The point of diminishing returns

happens when the concavity of P (t) changes, which is the when the derivative, P ′(t), switches from increasing to

decreasing. This happens when P ′(t) reaches its maximum at t = t0

10. (a) At t = 0, which corresponds to 1935, we have

P =1

1 + 3e−0.0275(0)= 0.25

showing that 25% of the land was in use in 1935.

(b) This model predicts that as t gets very large, P approaches 1. That is, the model predicts that in the long run, all the

land will be used for farming.

(c) To solve this graphically, enter the function into a graphing calculator and trace the resulting curve until it reaches

a height of 0.5, which occurs when t = 39.9 ≈ 40. Since t = 0 corresponds to 1935, t = 40 corresponds to

1935 + 40 = 1975. According to this model, the Tojolobal were using half their land in 1975. See Figure 4.125.

40 80 120 160 200

0.5

1

t (years since 1935)

P (proportion of land in use for farming)

Figure 4.125

(d) The point of diminishing returns occurs when P = L/2 or at one-half the carrying capacity. In this case, P = 1/2in 1975, as shown in part (c).

11. (a) Between day t = 0 and day t = 5,

Average

rate of change=

∆P

∆t=

222 − 95

5 − 0= 25.4 new cases per day.

Similarly, we find the values in Table 4.4.

Table 4.4

Days 0–5 5–12 12–19 19–26 26–33 33–40 40–47

New cases per day 25.4 35.4 47.1 44.0 35.7 24.1 13.4

Days 47–54 54–61 61–68 68–75 75–81 81–87

New cases per day 7.6 5.1 2.0 2.1 1.8 0.8

(b) The first computation showing a reduced rate of change was made with the data for t = 26, which corresponds to

April 12, 2003. On this date, epidemiologists could report that the rate of new cases had begun to slow.

4.7 SOLUTIONS 257

(c) The rate of change increases from t = 0 to t = 19, so the graph is concave up there. After t = 19, the rate of change

decreases and the graph is concave down. Exponential functions do not change concavity.

(d) The inflection point appears to occur near t = 19, when the average number of new cases per day stops increasing and

begins decreasing. At this point, we have P = 800, so we estimate the limiting value of P to be about 2 ·800 = 1600cases.

(e) The predicted limiting value is 1760, which is close to the actual value of 1755.

12. Substituting t = 0, 10, 20, . . . , 70 into the function P = 3.9(1.03)t gives the values in Table 4.5. Notice that the

agreement is very close, reflecting the fact that an exponential function models the growth well over the period 1790–

1860.

Table 4.5 Predicted versus actual US population

1790–1860, in millions. (exponential model)

Year Actual Predicted Year Actual Predicted

1790 3.9 3.9 1830 12.9 12.7

1800 5.3 5.2 1840 17.1 17.1

1810 7.2 7.0 1850 23.2 23.0

1820 9.6 9.5 1860 31.4 30.9

13. Substituting t = 0, 10, 20, 30, . . . into the function

P =187

1 + 47e−0.0318t

gives the values in Table 4.6. Notice that the agreement is close between the predicted and actual values.

Table 4.6

Year Actual Predicted Year Actual Predicted Year Actual Predicted

1790 3.9 3.9 1840 17.1 17.7 1890 62.9 63.3

1800 5.3 5.3 1850 23.1 23.5 1900 76.0 77.2

1810 7.2 7.2 1860 31.4 30.8 1910 92.0 91.9

1820 9.6 9.8 1870 38.6 39.9 1920 105.7 106.7

1830 12.9 13.2 1880 50.2 50.7 1930 122.8 120.8

1940 131.7 133.7

14. (a) We use k = 1.78 as a rough approximation. We let L = 5000 since the problem tells us that 5000 people eventually

get the virus. This means the limiting value is 5000.

(b) We know that

P (t) =5000

1 + Ce−1.78tand P (0) = 10

so

10 =5000

1 + Ce0=

5000

1 + C

10(1 + C) = 5000

1 + C = 500

C = 499.

(c) We have P (t) =5000

1 + 499e−1.78t. This function is graphed in Figure 4.126.


1 2 3 4 5 6 7

1000

2000

3000

4000 P (t) = 50001+499e−1.78t

week

number infected

Figure 4.126

(d) The point of diminishing returns appears to be at the point (3.5, 2500); that is, after 3 and a half weeks and when

2500 people are infected.

15. (a) The dose-response curve for this drug is shown if Figure 4.127.

150

100

R = 1001+100e−0.1x

x

Figure 4.127

(b) First solve for x:

R =100

1 + 100e−0.1x

R + 100Re−0.1x = 100

100Re−0.1x = 100 − R

e−0.1x =100 − R

100R

−0.1x = ln(

100 − R

100R

)

x = −10 ln(

100 − R

100R

)

.

Substituting the value of R = 50 we get

x = −10 ln(

100 − 50

100 · 50

)

≈ 46.0517019 ≈ 46.05

(c) Evaluate the formula for x obtained in part (b) for R = 20 and R = 70.

For R = 20,

x = −10 ln(

100 − 20

100 · 20

)

≈ 32.18875825 ≈ 32.12

For R = 70,

x = −10 ln(

100 − 70

100 · 70

)

≈ 54.5246805 ≈ 54.52

So the range of doses that is both safe and effective is between 32.12 mg and 54.52 mg.

4.7 SOLUTIONS 259

16. (a) The dose-response curve for product C crosses the minimum desired response line last, so it requires the largest dose

to achieve the desired response. The dose-response curve for product B crosses the minimum desired response line

first, so it requires the smallest dose to achieve the desired response.

(b) The dose-response curve for product A levels off at the highest point, so it has the largest maximum response. The

dose-response curve for product B levels off at the lowest point, so it has the smallest maximum response.

(c) Product C is the safest to administer because its slope in the safe and effective region is the least, so there is a broad

range of dosages for which the drug is both safe and effective.

17. (a) The fact that f ′(15) = 11 means that the slope of the curve at the inflection point, (15, 50) is 11. In terms of dose

and response, this large slope tells us that the range of doses for which this drug is both safe and effective is small.

(b) As we can see from Figure 4.77 in the text, a dose-response curve starts out concave up (slope increasing) and

switches to concave down (slope decreasing) at an inflection point. Since the slope at the inflection point (15, 50) is

11, and the slope is increasing before the inflection point, f ′(10) is less than 11. Since the slope is decreasing after

the inflection point, f ′(20) is also less than 11.

18. If the derivative of the dose-response curve is smaller, the slope is not as steep. Since the slope is not as steep, the

response increases less at the same dosage. Therefore, there is a wider range of dosages that are both safe and effective,

and consequently the dosage given to the patient does not have to be as exact.

19. The range of safe and effective doses begins at 10 mg where the drug is effective for 100 percent of patients. It ends at 18mg where the percent lethal curve begins to rise above zero.

20. When 50 mg of the drug is administered, it is effective for 85 percent of the patients and lethal for 6 percent.

21. (a) We have

L − P = L − L

1 + Ce−kt

=L(1 + Ce−kt) − L

1 + Ce−kt

=CLe−kt

1 + Ce−kt

= CPe−kt.

Dividing by P givesL − P

P= Ce−kt.

(b) The existing population is P . Since the carrying capacity is L, the additional population the environment can support

is L − P .

22. (a) We have

f(V ) =1

1 + e−(V +25)/2

=1

1 + e−25/2e−12

V.

Comparison with

f(V ) =L

1 + Ce−kV

shows that L = 1, k = 1/2 = 0.5, and C = e−25/2 = 3.7 · 10−6.

(b) Ten percent of the channels are open at potential V where

f(V ) = 0.1

1

1 + e−(V +25)/2= 0.1

V = −29.4 mV.

Half the channels are open when f(V ) = 0.5, or V = −25 mV.

Ninety per cent of the channels are open when f(V ) = 0.9, or V = −20.6 mV.




10 20

10

20

C = 12.4te−0.2t

t (hours)

C (ng/ml)

Figure 4.128

(b) The surge function y = atebt changes from increasing to decreasing at t = 1b

. For this function b = 0.2 so the peak

is at 10.2

= 5 hours. We can now substitute this into the formula to compute the peak concentration:

C = 12.4(5)e−0.2(5) = 22.8085254 ng/ml ≈ 22.8 ng/ml.

(c) Tracing along the graph of C = 12.4te−0.2t , we see it crosses the line C = 10 at t ≈ 1 hour and at t ≈ 14.4 hours.

Thus, the drug is effective for 1 ≤ t ≤ 14.4 hours.

(d) The drug drops below C = 4 for t > 20.8 hours. Thus, it is safe to take the other drug after 20.8 hours.

2.

−1 7

−0.4

0.4

t

C

Figure 4.129: a = 1

−1 7

−4

4

t

C

Figure 4.130: a = −10

−1 7

−0.04

0.04

t

C

Figure 4.131: a = 0.1

The parameter a apparently affects the height and direction of C = ate−bt. If a is positive, the “hump” is above the

t-axis. If a is negative, it’s below the t-axis. If a > 0, the larger the value of a, the larger the maximum value of C. If

a < 0, the more negative the value of a, the smaller the minimum value of C.

3. Newborn babies are generally not as effective as adults in eliminating drugs from the body. See the figure in the text. The

level of the drug in the body begins to decline after two hours in an adult, but not until six hours have elapsed in a newborn.

In addition, the concentration in the blood reaches a higher level in the newborn than it does in the adult. Graphs such as

this explain why physicians modify the doses for drugs given to pregnant women and women who are breastfeeding.

4.

10 20

10

20

C = 17.2te−0.4t

t (hours)

C (ng/ml)

Figure 4.132

(a) See Figure 4.132. The first dose becomes ineffective when

10 = 17.2te−0.4t.

4.8 SOLUTIONS 261

To find this value of t, trace along the curve C = 17.2te−0.4t , giving t ≈ 6 hours. Thus, the second dose should be

given after about 6 hours.

(b) By tracing, we find that the second dose becomes effective about 1 hour after it is given. We want it to become

effective when t ≈ 6, so the second dose should be given at t ≈ 5 hours.

5. Large quantities of water dramatically increase the value of the peak concentration, but do not change the amount of time it

takes to reach the peak concentration. The effect of the volume of water taken with the drug wears off after approximately

6 hours.

6. Figure 4.133 has its maximum at t = 1.3 hours, C = 23.6 ng/ml.

1 2 3 4 5 6 7

5

10

15

20

25

t (hours)

Concentration (ng/ml)

Figure 4.133

7. For cigarettes the peak concentration is approximately 17 ng/ml, the time until peak concentration is about 10 minutes,

and the nicotine is at first eliminated quickly, but after about 45 minutes it is eliminated more slowly.

For chewing tobacco, the peak concentration is approximately 14 ng/ml and the time until peak concentration is

about 30 minutes (when chewing stops). The nicotine is eliminated at a slow, somewhat erratic rate.

For nicotine gum the peak concentration is about 10 ng/ml and the time until peak concentration is approximately

45 minutes. The nicotine is eliminated at a very slow but steady rate.

8. (a) Differentiating using the product rule gives

C′(t) = 20e−0.03t + 20t(−0.03)e−0.03t

= 20(1 − 0.03t)e−0.03t.

At the peak concentration, C′(t) = 0, so

20(1 − 0.03t)e−0.03t = 0

t =1

0.03= 33.3 minutes.

When t = 33.3, the concentration is

C = 20(33.3)e−0.03(33.3) ≈ 245 ng/ml.

See Figure 4.134. The curve peaks after 33.3 minutes with a concentration of 244.9 ng/ml.

50 100 150 200 250

50

100

150

200

minutes

concentration (ng/ml)

C(t) = 20te−0.03t

Figure 4.134


(b) After 15 minutes, the drug concentration will be C(15) = 20(15)e(−0.03)(15) ≈ 191 ng/ml. After an hour, the

concentration will be C(60) = 20(60)e−0.03(60) ≈ 198 ng/ml.

(c) We want to know where C(t) = 10. We estimate from the graph. It looks like C(t) = 10 after 190 minutes or a little

over 3 hours.

9. We find values of the parameters in the function C = ate−bt to create a local maximum at the point (1.3, 23.6). We first

set the derivative equal to zero and solve for t to find critical points. Using the product rule, we have:

dC

dt= at(e−bt(−b)) + a(e−bt) = 0

ae−bt(−bt + 1) = 0

t =1

b.

The only critical point is at t = 1/b. Since we want a critical point at t = 1.3, we substitute and solve for b:

1.3 =1

b

b =1

1.3= 0.769.

To find the value of a, we use the fact that C = 23.6 when t = 1.3. We have:

a(1.3)e−0.769(1.3) = 23.6

a · 1.3e−1 = 23.6

a =23.6

1.3e−1= 49.3.

We have a = 49.3 and b = 0.769.

10. (a) The IV method reaches peak concentration the fastest, it in fact begins at its peak.

The P-IM method reaches peak concentration the slowest.

(b) The IV method has the largest peak concentration.

The PO method has the smallest peak concentration.

(c) The IV method wears off the fastest.

The P-IM method wears off the slowest.

(d) The P-IM method has the longest effective duration.

The IV method has the shortest effective duration.

(e) It is effective for approximately 5 hours.

11. Food dramatically increases the value of the peak concentration but does not affect the time it takes to reach the peak

concentration. The effect of food is stronger during the first 8 hours.

12. (a) The value f(t) = e−t − e−2t is the vertical distance between the two curves in Figure 4.135. That distance is 0 at

t = 0. The distance first increases as t increases from 0. As t gets larger and larger, the distance eventually begins to

decrease toward 0 because both e−t and e−2t approach the same horizontal asymptote, the t-axis. Thus the graph of

f(t) shows a rapid increase from zero followed by a slow decrease back toward zero; this is the shape of a surge.

(b) At a critical point we have

f ′(t) = −e−t + 2e−2t = 0.

Hence

e−t = 2e−2t

et = 2

t = ln 2 = 0.693.

At the critical point, f(ln 2) = 1/4 = 0.250. The critical point is (ln 2, 1/4) = (0.693, 0.250).

At an inflection point we have

f ′′(t) = e−t − 4e−2t = 0.

Hence

e−t = 4e−2t

et = 4

t = ln 4 = 1.386

f(ln 4) =3

16= 0.1875.

SOLUTIONS to Review Problems For Chapter Four 263

The inflection point is (2 ln 2, 3/16) = (1.386, 0.187). See Figure 4.136

1 2 3 4 5

0.5

1

e−2t

e−t

t

y

Figure 4.135

1 2 3 4 5

0.1

0.2

0.3(log 2, 1/4)

(log 4, 3/16)

e−t − e−2t

x

y

Figure 4.136

13. (a) Products A and B have much higher peak concentrations than products C and D. Product A reaches its peak con-

centration slightly before products B, C, and D, which all take about the same time to reach peak concentration.

(b) If the minimum effective concentration were low, perhaps 0.2, and the maximum safe concentration were also low,

perhaps 1.0, then products C and D would be the preferred drugs since they do not enter the unsafe range while

being well within the effective range.

(c) If the minimum effective concentration were high, perhaps 1.2, and the maximum safe concentration were also high,

perhaps 2.0, then product A would be the preferred drug since it does not enter the unsafe range and it is the only

drug that is in the effective range for a substantial amount of time.

14. In general, a faster dissolution rate corresponds to a larger peak concentration. Dissolution rate does not affect time to

reach peak concentration, as the four products have different dissolution rates but virtually the same time to reach peak

concentration.

Solutions for Chapter 4 Review


1 2 3 4 5

2

4

6

8

Global and local min

Global and local max

Local min

x

y

Figure 4.137

2. The global maximum is achieved at the two local maxima, which are at the same height. See Figure 4.138.

2 4 6 8 10

1

2

3

4

Local min

Local andglobal max


Local andglobal max

Local min

x

y

Figure 4.138


3. (a) We have f ′(x) = 10x9 − 10 = 10(x9 − 1). This is zero when x = 1, so x = 1 is a critical point of f . For values of

x less than 1, x9 is less than 1, and thus f ′(x) is negative when x < 1. Similarly, f ′(x) is positive for x > 1. Thus

f(1) = −9 is a local minimum.

We also consider the endpoints f(0) = 0 and f(2) = 1004. Since f ′(0) < 0 and f ′(2) > 0, we see x = 0 and

x = 2 are local maxima.

(b) Comparing values of f shows that the global minimum is at x = 1, and the global maximum is at x = 2.

4. (a) f ′(x) = 1 − 1/x. This is zero only when x = 1. Now f ′(x) is positive when 1 < x ≤ 2, and negative when

0.1 < x < 1. Thus f(1) = 1 is a local minimum. The endpoints f(0.1) ≈ 2.4026 and f(2) ≈ 1.3069 are local

maxima.

(b) Comparing values of f shows that x = 0.1 gives the global maximum and x = 1 gives the global minimum.

5. We rewrite h(z) as h(z) = z−1 + 4z2.

Differentiating gives

h′(z) = −z−2 + 8z,


−z−2 + 8z = 0

z−2 = 8z

8z3 = 1

z3 =1

8

z =1

2.

Since h′ is negative for 0 < z < 1/2 and h′ is positive for z > 1/2, there is a local minimum at z = 1/2.

Since h(z) → ∞ as z → 0+ and as z → ∞, the local minimum at z = 1/2 is a global minimum; there is no global

maximum. See Figure 4.139. Thus, the global minimum is h(1/2) = 3.

1/2

3h(z) = 1

z+ 4z2

z

Figure 4.139

6. Since g(t) is always decreasing for t ≥ 0, we expect it to a global maximum at t = 0 but no global minimum. At t = 0,

we have g(0) = 1, and as t → ∞, we have g(t) → 0.

Alternatively, rewriting as g(t) = (t3 + 1)−1 and differentiating using the chain rule gives

g′(t) = −(t3 + 1)−2 · 3t2.

Since 3t2 = 0 when t = 0, there is a critical point at t = 0, and g decreases for all t > 0. See Figure 4.140.

1

g(t) =1

t3 + 1

x

Figure 4.140


7. We begin by rewriting f(x):

f(x) =1

(x − 1)2 + 2= ((x − 1)2 + 2)−1 = (x2 − 2x + 3)−1.

Differentiating using the chain rule gives

f ′(x) = −(x2 − 2x + 3)−2(2x − 2) =2 − 2x

(x2 − 2x + 3)2,


2 − 2x

(x2 − 2x + 3)2= 0

2 − 2x = 0

2x = 2

x = 1.

Since f ′ is positive for x < 1 and f ′ is negative for x > 1, there is a local maximum at x = 1.

Since f(x) → 0 as x → ∞ and as x → −∞, the local maximum at x = 1 is a global maximum; there is no global

minimum. See Figure 4.141. Thus, the global maximum is f(1) = 1/2.

1

1/2f(x) =

1

(x − 1)2 + 2

x

Figure 4.141

8. There are several possibilities. The price could have been increasing during the last few days of June, reaching a high

point on July 1, then going back down during the first few days of July. In this case there was a local maximum in the

price on July 1.

The price could have been decreasing during the last few days of June, reaching a low point on July 1, then going

back up during the first few days of July. In this case there was a local minimum in the price on July 1.

It is also possible that there was neither a local maximum nor a local minimum in the price on July 1. This could

have happened two ways. On the one hand, the price could have been rising in late June, then held steady with no change

around July 1, after which the price increased some more. On the other hand, the price could have been falling in late

June, then held steady with no change around July 1, after which the price fell some more. The key feature in these critical

point scenarios is that there was no appreciable change in the price of the stock around July 1.

9. (a) Varying the a parameter changes the peak concentration proportionally, but does not change the time to reach peak

concentration .

(b) The value of the peak concentration decreases as b increases and the time until peak concentration decreases as bincreases.

(c) As a gets larger, the time until peak concentration decreases because it is only affected by the constant in the exponent.

The value of the peak concentration is not affected as a changes because the peak always occurs at t = 1/b and if

a = b, the value of the peak concentration is c = a · 1ae−a·1/a = e−1, independent of a.

10. (a) Increasing for x > 0, decreasing for x < 0.

(b) f(0) is a local and global minimum, and f has no global maximum.

11. (a) Increasing for all x.

(b) No maxima or minima.

12. (a) Decreasing for x < 0, increasing for 0 < x < 4, and decreasing for x > 4.

(b) f(0) is a local minimum, and f(4) is a local maximum.

13. (a) Decreasing for x < −1, increasing for −1 < x < 0, decreasing for 0 < x < 1, and increasing for x > 1.

(b) f(−1) and f(1) are local minima, f(0) is a local maximum.


14. (a) We know that Profit = Revenue − Cost, so differentiating with respect to q gives:

Marginal Profit = Marginal Revenue − Marginal Cost.

We see from the figure in the problem that just to the left of q = a, marginal revenue is less than marginal cost, so

marginal profit is negative there. To the right of q = a marginal revenue is greater than marginal cost, so marginal

profit is positive there. At q = a marginal profit changes from negative to positive. This means that profit is decreasing

to the left of a and increasing to the right. The point q = a corresponds to a local minimum of profit, and does not

maximize profit. It would be a terrible idea for the company to set its production level at q = a.

(b) We see from the figure in the problem that just to the left of q = b marginal revenue is greater than marginal cost, so

marginal profit is positive there. Just to the right of q = b marginal revenue is less than marginal cost, so marginal

profit is negative there. At q = b marginal profit changes from positive to negative. This means that profit is increasing

to the left of b and decreasing to the right. The point q = b corresponds to a local maximum of profit. In fact, since

the area between the MC and MR curves in the figure in the text between q = a and q = b is bigger than the area

between q = 0 and q = a, q = b is in fact a global maximum.

15. (a) Centimeters per week.

(b) At week 24 the fetus is growing at a rate of 1.6 cm/week.

16. (a) The tangent line to the graph is steeper at 20 weeks then at 36 weeks, so f ′(20) is greater than f ′(36).

(b) The fetus increases its length more rapidly at week 20 than at week 36.

17. (a) The inflection point occurs at about week 14 where the graph changes from concave up to concave down.

(b) The fetus increases its length faster at week 14 than at any other time during its gestation.

18. We estimate the derivatives at 20 and 36 weeks by drawing tangent lines to the length graph, shown in Figure 4.142, and

calculating their slopes.

(a) Two points on the tangent line at 20 weeks are (6, 0) and (32, 49). Thus,

f ′(20) =49 − 0

32 − 6= 1.9 cm/week.

(b) Two points on the tangent line at 36 weeks are (0, 20) and (40, 50). Thus

f ′(36) =50 − 20

40 − 0= 0.75 cm/week.

(c) The average rate of growth is the slope of the secant line from (0, 0) to (40, 50). Thus

Average rate of change =50 − 0

40 − 0= 1.25 cm/week.

4 8 12 16 20 24 28 32 36 40

10

20

30

40

50

60

?

Slope = f ′(36)

� Slope = f ′(20)

t, age of fetus (weeks after last menstruation)

length (centimeters)

Figure 4.142

19. Using the product rule on the function f(x) = axebx, we have f ′(x) = aebx + abxebx = aebx(1 + bx). We want

f( 13) = 1, and since this is to be a maximum, we require f ′( 1

3) = 0. These conditions give

f(1/3) = a(1/3)eb/3 = 1,

f ′(1/3) = aeb/3(1 + b/3) = 0.


Since ae(1/3)b is non-zero, we can divide both sides of the second equation by ae(1/3)b to obtain 0 = 1 + b3

. This

implies b = −3. Plugging b = −3 into the first equation gives us a( 13)e−1 = 1, or a = 3e. How do we know we have

a maximum at x = 13

and not a minimum? Since f ′(x) = aebx(1 + bx) = (3e)e−3x(1 − 3x), and (3e)e−3x is always

positive, it follows that f ′(x) > 0 when x < 13

and f ′(x) < 0 when x > 13

. Since f ′ is positive to the left of x = 13

and

negative to the right of x = 13

, f( 13) is a local maximum.

20. Local maximum for some θ, with 1.1 < θ < 1.2, since f ′(1.1) > 0 and f ′(1.2) < 0.

Local minimum for some θ, with 1.5 < θ < 1.6, since f ′(1.5) < 0 and f ′(1.6) > 0.

Local maximum for some θ, with 2.0 < θ < 2.1, since f ′(2.0) > 0 and f ′(2.1) < 0.

21. (a) The derivative of f(x) = x5 + x + 7 is f ′(x) = 5x4 + 1. The derivative is always positive.

(b) Since f ′(x) 6= 0 for any value of x, there are no critical points for the function. Since f ′(x) is positive for all x, the

function is increasing for all x, it crosses the x-axis at most once. Since f(x) → +∞ as x → +∞ and f(x) → −∞as x → −∞, the graph of f crosses the x-axis once. So we conclude that f(x) has one real root.

22. (a)

x1 x3x5

x2

x4

f ′(x)

x

(b) f ′(x) changes sign at x1, x3, and x5.

(c) f ′(x) has local extrema at x2 and x4.

23. The local maxima and minima of f correspond to places where f ′ is zero and changes sign or, possibly, to the endpoints

of intervals in the domain of f . The points at which f changes concavity correspond to local maxima and minima of f ′.

The change of sign of f ′, from positive to negative corresponds to a maximum of f and change of sign of f ′ from negative

to positive corresponds to a minimum of f .

24. Since the function is positive, the graph lies above the x-axis. If there is a global maximum at x = 3, t′(x) must be

positive, then negative. Since t′(x) and t′′(x) have the same sign for x < 3, they must both be positive, and thus the

graph must be increasing and concave up. Since t′(x) and t′′(x) have opposite signs for x > 3 and t′(x) is negative,

t′′(x) must again be positive and the graph must be decreasing and concave up. A possible sketch of y = t(x) is shown

in Figure 4.143.

(3, 3)

y = t(x)

y

x

Figure 4.143

25. (a) Quadratic polynomial (degree 2) with negative leading coefficient.

(b) Exponential.

(c) Logistic.

(d) Logarithmic.

(e) This is a quadratic polynomial (degree 2) and positive leading coefficient.

(f) Exponential.

(g) Surge

26. First find the marginal revenue and marginal cost. Note that each product sells for $788, so revenue is given by R(q) =788q.

MR = R′(q) = 788,

MC = C′(q) = 3q2 − 120q + 1400.


Setting MR = MC yields

3q2 − 120q + 1400 = 788

3q2 − 120q + 612 = 0.

This factors to

3(q − 34)(q − 6) = 0,

so MR = MC at q = 34 and q = 6. We now find the profit at these points:

R(6) − C(6) = 788 · 6 −[

63 − 60 · 62 + 1400 · 6 + 1000]

= −$2728.

R(34) − C(34) = 788 · 34 −[

343 − 60 · 342 + 1400 · 34 + 1000]

= $8248

We must also try the endpoints

R(0) − C(0) = 788 · 0 −[

03 − 60 · 02 + 1400 · 0 + 1000]

= −$1000

R(50) − C(50) = 788 · 50 −[

503 − 60 · 502 + 1400 · 50 + 1000]

= −$6600.

From this we see that profit is maximized at q = 34 units. The total cost at q = 34 is C(34) = $18,544. The total revenue

at q = 34 is R(34) = $26,792 and

Total profit = 26,792 − 18,544 = $8248.

34

20,000

40,000

6

?Cost

6?Profit 6

?

Revenue

C(q)

R(q)

q

$

Figure 4.144

27. (a) The fixed cost is 0 because C(0) = 0.

(b) Profit, π(q), is equal to money from sales, 7q, minus total cost to produce those items, C(q).

π = 7q − 0.01q3 + 0.6q2 − 13q

π′ = −0.03q2 + 1.2q − 6

π′ = 0 if q =−1.2 ±

√

(1.2)2 − 4(0.03)(6)

−0.06≈ 5.9 or 34.1.

Now π′′ = −0.06q + 1.2, so π′′(5.9) > 0 and π′′(34.1) < 0. This means q = 5.9 is a local min and q = 34.1 a

local max. We now evaluate the endpoint, π(0) = 0, and the points nearest q = 34.1 with integer q-values:

π(35) = 7(35) − 0.01(35)3 + 0.6(35)2 − 13(35) = 245 − 148.75 = 96.25,

π(34) = 7(34) − 0.01(34)3 + 0.6(34)2 − 13(34) = 238 − 141.44 = 96.56.

So the (global) maximum profit is π(34) = 96.56. The money from sales is $238, the cost to produce the items is

$141.44, resulting in a profit of $96.56.

(c) The money from sales is equal to price×quantity sold. If the price is raised from $7 by $x to $(7 + x), the result is a

reduction in sales from 34 items to (34 − 2x) items. So the result of raising the price by $x is to change the money

from sales from (7)(34) to (7 + x)(34 − 2x) dollars. If the production level is fixed at 34, then the production costs

are fixed at $141.44, as found in part (b), and the profit is given by:

π(x) = (7 + x)(34 − 2x) − 141.44


This expression gives the profit as a function of change in price x, rather than as a function of quantity as in part (b).

We set the derivative of π with respect to x equal to zero to find the change in price that maximizes the profit:

dπ

dx= (1)(34 − 2x) + (7 + x)(−2) = 20 − 4x = 0

So x = 5, and this must give a maximum for π(x) since the graph of π is a parabola which opens downward. The

profit when the price is $12 (= 7 + x = 7 + 5) is thus π(5) = (7 + 5)(34 − 2(5)) − 141.44 = $146.56. This is

indeed higher than the profit when the price is $7, so the smart thing to do is to raise the price by $5.

28. Note that profit = π(q) = R(q)−C(q), so to find profit, we must find an expression for R(q). If p is the price of a single

item,

R(q) = p · q.Substituting p = b1 − a1q gives

R(q) = (b1 − a1q) · q= b1q − a1q

2and

π(q) = R(q) − C(q)

= b1q − a1q2 − b2 − a2q.

Finding the derivative and setting it equal to 0 yields

π′(q) = b1 − 2a1q − a2 = 0

so q =−a2 + b1

2a1

is a critical point for π(q). Using the second derivative test, we see π is concave down,

π′′(q) = −2a1

Since a1 is positive, −2a1 will always be negative, so, since π′′(q) < 0 for all q, q = (−a2 + b1)/(2a1) is a local

maximum. This q is the only critical point, so it is the global maximum for π(q).

29. (a) π(q) is maximized when R(q) > C(q) and they are as far apart as possible. See Figure 4.145.

(b) π′(q0) = R′(q0) − C′(q0) = 0 implies that C′(q0) = R′(q0) = p.

Graphically, the slopes of the two curves at q0 are equal. This is plausible because if C′(q0) were greater than por less than p, the maximum of π(q) would be to the left or right of q0, respectively. In economic terms, if the cost

were rising more quickly than revenues, the profit would be maximized at a lower quantity (and if the cost were rising

more slowly, at a higher quantity).


q0

C(q)

R(q)

Imaximum π(q)

q

$

Figure 4.145

q0

p

q

C′(q)$

Figure 4.146


30. (a) We know that the average cost is given by

a(q) =C(q)

q.

Thus the average cost is

a(q) = 0.04q2 − 3q + 75 +96

q.

(b) Average cost is graphed in Figure 4.147.

20 40 60 80 100

50

100

150

q

a

a(q)

Figure 4.147

(c) Looking at the graph of the average value we see that it is decreasing for q < 38. Likewise we see that a(q) is

increasing for q > 38.

(d) Thus the average cost hits its minimum at about q = 38 with the value of about $21.

31. If the minimum average cost occurs at a production level of 15,000 units, the line from the origin to the curve is tangent

to the curve at that point. The slope of this line is 25, so the cost of producing 15,000 units is 25(15,000) = 375,000. See

Figure 4.148.

15,000

375,000

Quantity

Cost

Figure 4.148

32. The point F representing the fixed costs is the vertical intercept of the cost function, C(q).

The break-even level of production is where R(q) = C(q), so the point B is where the two graphs intersect. There

are two such points in Figure 4.149. We have marked the one at which production first becomes profitable.

The marginal cost is C′(q), so the point M is where the slope of C(q) is minimum.

The average cost is minimized when a(q) = C′(q). Since a(q) equals the slope of a line joining the origin to a point

(q, C(q)) on the graph of C(q), the point A is where such a line is tangent to the graph of C(q).

The profit is P (q) = R(q)−C(q), so the point P occurs where the two graphs are furthest apart and R(q) > C(q).

See Figure 4.149.


B M A PF

C(q)

R(q)

q

$

Figure 4.149

33. C′(q) decreases with q because C(q) is concave down. Therefore, C′(2) is larger then C′(3).

34. Note that (C(5) − C(3))/(5 − 3) is the slope of the secant line between q = 3 and q = 5. The value of (C(5) −C(3))/(5−3) is larger because the slope of the secant line is larger than the slope of the tangent line at its right endpoint.

35. Both quantities are slopes of secant lines with left endpoint at q = 50. The value of (C(75) − C(50))/25 is larger as the

slope of the secant line decreases as the right endpoint moves to the right.

36. Both quantities are slopes of secant lines. Since the secant line between q = 25 and q = 75 is to the left of the secant line

between q = 50 and q = 100, and C(q) is concave down, (C(75) − C(25))/50 is larger.

37. Since C′(3) is the slope at q = 3, and C(3)/3 is the slope of the line from the origin to the point q = 3 on the curve

C(q), the value of C(3)/3 is larger.

38. Since C(q)/q is the slope of the line from the origin to the point q on the curve C(q), the concavity of C(q) tells us that

the value of C(10)/10 is larger.

39. E =

∣

∣

∣

∣

p

q

dq

dp

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· d

dp(2000 − 5p)

∣

∣

∣

∣

=

∣

∣

∣

∣

p

q· (−5)

∣

∣

∣

∣

, so E =5p

q. At a price of $20.00, the number of items produced is

q = 2000 − 5(20) = 1900, so at p = 20, we have

E =5(20)

(1900)≈ 0.05.

Since 0 ≤ E < 1, this product has inelastic demand–a 1% change in price will only decrease demand by 0.05%.

40. Since R = pq, we have dR/dp = p(dq/dp) + q. We are assuming that

E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

> 1

so, removing the absolute values

−E =p

q

dq

dp< −1.

Multiplication by q gives

pdq

dp< −q

and hencedR

dp= p

dq

dp+ q < 0

41. Since R = pq, we have dR/dp = p(dq/dp) + q. We are assuming that

0 ≤ E =

∣

∣

∣

∣

p

q· dq

dp

∣

∣

∣

∣

< 1

so, removing the absolute values

0 ≥ −E =p

q

dq

dp> −1

Multiplication by q gives

pdq

dp> −q


and hencedR

dp= p

dq

dp+ q > 0

42. (a) If we graph the data, we see that it looks like logistic growth. (See Figure 4.150.) But logistic growth also makes

sense from a common-sense viewpoint. As cable television “catches on,” the percentage of households which have it

will at first grow exponentially but then slow down after more and more people have it. Eventually, nearly everyone

who will subscribe already has and the percentage levels off.

1975 1980 1985 1990 1995 2000 20050

25

50

75

100

year

% of households with cable

Figure 4.150

(b) The point of diminishing returns happens around 35%. This predicts a carrying capacity of 70% which is pretty close

to the 68.9% we see in 2002 and 68.0% we see in 2003.

(c) 68.8%(d) The limiting value predicts the percentage of households that will eventually have cable. This model predicts that

there will never be a time when more than 68.8% of households have cable.

43. Let l and w be the length and width of the rectangle. Since the perimeter is 200 meters, we have

2w + 2l = 200

l = 100 − w.

The area is

A(w) = w · l = w(100 − w) = 100w − w2.

At the critical points

A′(w) = 100 − 2w = 0,

so

w = 50, l = 50.

The rectangle of perimeter 200 meters with maximum area is the 50 meter by 50 meter square. Its area is w · l = 50 ·50 =2500 meters2.

44. The volume is given by V = x2y. The surface area is given by

S = 2x2 + 4xy

= 2x2 + 4xV/x2 = 2x2 + 4V/x.

To find the dimensions which minimize the area, find x such that dS/dx = 0:

dS

dx= 4x − 4V

x2= 0

x3 = V.

Solving for x gives x = 3√

V = y. To see that this gives a minimum, note that for small x, S ≈ 4V/x is decreasing. For

large x, S ≈ 2x2 is increasing. Since there is only one critical point, it must give a global minimum. Therefore, when the

width equals the height, the surface area is minimized.


45. (a) Setting the derivative of p(1 − p)4 equal to 0

d

dp(p(1− p)4) = (1 − p)4 − 4p(1 − p)3 = 0

(1 − p)3(1 − p − 4p) = 0

(1 − p)3(1 − 5p) = 0

p = 1/5, 1.

Thus, the critical points are p = 1/5 and p = 1.

(b) Since

d2

dp2(p(1− p)4) =

d

dp((1 − p)4 − 4p(1 − p)3)

= −4(1 − p)3 − 4(1 − p)3 + 12p(1 − p)2

= 4(1 − p)2(−2(1 − p) + 3p)

= 4(1 − p)2(−2 + 5p),

substituting p = 1/5 and p = 1, we have

d2

dp2(p(1 − p)4)

∣

∣

∣

∣

∣

p=1/5

= 4(

4

5

)2

(−1) < 0 andd2

dp2(p(1 − p)4)

∣

∣

∣

∣

∣

p=1

= 0

Thus p = 1/5 is a local maximum. The second derivative test does not enable us to classify p = 1. However,

p(1 − p)4 is positive everywhere except at p = 0 and p = 1, where it is 0. Thus, p = 1 is a local minimum.

(c) The global maximum occurs at the local maximum, at p = 1/5, so

Maximum =1

5

(

1 − 1

5

)4

=44

55=

256

3125.

The global minimum occurs at the end points, so

Minimum = 0(1 − 0)4 = 1(1 − 1)4 = 0.

46. The triangle in Figure 4.151 has area, A, given by

A =1

2xy =

x

2e−x/3.

At a critical point,

dA

dx=

1

2e−x/3 − x

6e−x/3 = 0

1

6e−x/3(3 − x) = 0

x = 3.

Substituting the critical point and the endpoints into the formula for the area gives:

For x = 1, we have A = 12e−1/3 = 0.358

For x = 3, we have A = 32e−1 = 0.552

For x = 5, we have A = 52e−5/3 = 0.472

Thus, the maximum area is 0.552 and the minimum area is 0.358.

-� x

6?y

(x, e−x/3)

x

y

Figure 4.151


47. (a) The distance the pigeon flies over water is

BP =AB

sin θ=

500

sin θ,

and over land is

PL = AL − AP = 2000 − 500

tan θ= 2000 − 500 cos θ

sin θ.

Therefore the energy required is

E = 2e(

500

sin θ

)

+ e(

2000 − 500 cos θ

sin θ

)

= 500e(

2 − cos θ

sin θ

)

+ 2000e, for arctan(

500

2000

)

≤ θ ≤ π

2.

(b) Notice that E and the function f(θ) =2 − cos θ

sin θmust have the same critical points since the graph of E is just a

stretch and a vertical shift of the graph of f . The graph of2 − cos θ

sin θfor arctan( 500

2000) ≤ θ ≤ π

2in Figure 4.152

shows that E has precisely one critical point, and that a minimum for E occurs at this point.

arctan 5002000

π2

π3

θ

Figure 4.152: Graph of f(θ) = 2−cos θsin θ

for arctan( 5002000

) ≤ θ ≤ π2

To find the critical point θ, we solve f ′(θ) = 0 or

E′ = 0 = 500e

(

sin θ · sin θ − (2 − cos θ) · cos θ

sin2 θ

)

= 500e(

1 − 2 cos θ

sin2 θ

)

.

Therefore 1 − 2 cos θ = 0 and so θ = π/3.

(c) Letting a = AB and b = AL, our formula for E becomes

E = 2e(

a

sin θ

)

+ e(

b − a cos θ

sin θ

)

= ea(

2 − cos θ

sin θ

)

+ eb, for arctan(

a

b

)

≤ θ ≤ π

2.

Again, the graph of E is just a stretch and a vertical shift of the graph of2 − cos θ

sin θ. Thus, the critical point θ = π/3

is independent of e, a, and b. But the maximum of E on the domain arctan(a/b) ≤ θ ≤ π2

is dependent on the ratio

a/b =AB

AL. In other words, the optimal angle is θ = π/3 provided arctan(a/b) ≤ π

3; otherwise, the optimal angle

is arctan(a/b), which means the pigeon should fly over the lake for the entire trip—this occurs when a/b > 1.733.


48. (a) Differentiating using the chain rule gives

p′(x) =1

σ√

2πe−(x−µ)2/(2σ2) ·

(

−2(x − µ)

2σ2

)

= − (x − µ)

σ3√

2πe−(x−µ)2/(2σ2).

So p′(x) = 0 where x − µ = 0, so x = µ.

Differentiating again using the product rule gives

p′′(x) =−1

σ3√

2π

(

1 · e−(x−µ)2/(2σ2) + (x − µ)e−(x−µ)2/(2σ2) ·(

−2(x − µ)

2σ2

))

= −e−(x−µ)2/(2σ2)

σ5√

2π(σ2 − (x − µ)2).

Substituting x = µ gives

p′′(µ) = − e0

σ5√

2πσ2 = − 1

σ3√

2π.

Since p′′(µ) < 0, there is a local maximum at x = µ. Since this local maximum is the only critical point, it is a

global maximum. See Figure 4.153.

(b) Using the formula for p′′(x), we see that p′′(x) = 0 where

σ2 − (x − µ)2 = 0

x − µ = ±σ

x = µ ± σ.

Since p′′(x) changes sign at x = µ + σ and x = µ − σ, these are both points of inflection. See Figure 4.153.

µµ − σ µ + σ

1

x

Inflectionpoint

Maximum

Inflectionpoint

p(x)

- �

Figure 4.153

49. (a) Before differentiating, we multiply out, giving

dP

dt= kP

(

1 − P

L

)

= kP − kP 2

L.

Since k and L are constant, the chain rule gives

d2P

dt2=

d

dt

(

dP

dt

)

=d

dt

(

kP − kP 2

L

)

= kdP

dt− 2kP

L

dP

dt.

(b) Setting d2P/dt2 = 0 gives

0 = kdP

dt− 2kP

L

dP

dt=

kdP

dt

(

1 − 2P

L

)

.

So

1 − 2P

L= 0,

that is, when

P =L

2,

we have d2P/dt2 = 0; that is, we are at the point of diminishing returns.


50. When bemetizide is taken with triamterene, rather than by itself, the peak concentration is lower, the time it takes to reach

peak concentration is about the same, the time until onset of effectiveness is slightly less, and the duration of effectiveness

is less. If bemetizide became unsafe in a concentration greater than 70 ng/ml, then it would be wise to take it with

triamterene.

51. To find the intercepts of f(x), we first find the y-intercept, which occurs at f(0) = sin(02) = 0. To find the x-intercepts

on the given interval, we use a calculator to graph f(x) as shown in Figure 4.154:

x

f(x) = sin(x2)

Figure 4.154

x

f ′(x)

Figure 4.155

x

f ′′(x)

Figure 4.156

We can use a calculator’s root-finding capability to find where f(x) = 0. We get:

x = 0, and x = 1.77, and x = 2.51,

so our intercepts are

(0, 0), (1.77, 0), (2.51, 0).

To find critical points, we look for where f ′(x) = 0 or f ′ is undefined. Using a calculator’s differentiation and graphing

features provides a graph of f ′(x) shown in Figure 4.155.

Since in Figure 4.155 all the critical points (x, f(x)) have f ′(x) = 0, we use the calculator’s root-finding capability

to find the critical points where f ′(x) = 0:

x = 0, x = 1.25, x = 2.17, and x = 2.80.

Writing these out as coordinates, the critical points are at

(0, 0), (1.25, 1), (2.17,−1), (2.80, 1).

Similarly, inflection points occur where f ′′(x) changes from negative to positive or vice versa. We can look for such

points on a graph of f ′′(x), shown in Figure 4.156.

We can use the calculator’s root-finding capability on the f ′′(x) to get these inflection points:

x = 0.81, and x = 1.81, and x = 2.52

The inflection points have coordinates (0.81, f(0.81)), etcetera and so they are

(0.81, 0.61), (1.81,−0.13), (2.52, 0.07).

Notice that the intercepts can also be computed algebraically, since sin(x2) = 0 when x2 = 0, π, 2π. The solutions

are x = 0, x =√

π = 1.77, x =√

2π = 2.51. Similarly, if we set f ′(x) = 2x cos(x2) = 0, we get x = 0

or x2 = π/2, 3π/2, 5π/2. Thus, the critical points are x = 0, x =√

π/2 = 1.25, x =√

3π/2 = 2.17, and

x =√

5π/2 = 2.80.


−10 10

−10

10a = 0.5

a = 3

x

Figure 4.157

(b) The function f(x) = x+a sin x is increasing for all x if f ′(x) > 0 for all x. We have f ′(x) = 1+a cos x. Because

cos x varies between −1 and 1, we have 1 + a cos x > 0 for all x if −1 < a < 1 but not otherwise. When a = 1,

the function f(x) = x + sin x is increasing for all x, as is f(x) = x − sin x, obtained when a = −1. Thus f(x) is

increasing for all x if −1 ≤ a ≤ 1.



−10 10

−100

100a = 1

a = 20

x

Figure 4.158

(b) The function f(x) = x2 + a sin x is concave up for all x if f ′′(x) > 0 for all x. We have f ′′(x) = 2 − a sin x.

Because sin x varies between −1 and 1, we have 2 − a sin x > 0 for all x if −2 < a < 2 but not otherwise. Thus

f(x) is concave up for all x if −2 < a < 2.

54. (a) For a point (t, s), the line from the origin has rise = s and run = t; See Figure 4.159. Thus, the slope of the line OPis s/t.

(b) Sketching several lines from the origin to points on the curve (see Figure 4.160), we see that the maximum slope

occurs at the point P , where the line to the origin is tangent to the graph. Reading from the graph, we see t ≈ 2 hours

at this point.

O

(t, s)

P

-� t

6

?

s

t (hours)

s (km)

Figure 4.159

P

t (hours)

s (km)

Figure 4.160

(c) The instantaneous speed of the cyclist at any time is given by the slope of the corresponding point on the curve. At

the point P , the line from the origin is tangent to the curve, so the quantity s/t equals the cyclist’s speed at the point

P .

55. (a) The line in Figure 4.161 has slope equal to the rate worms arrive. To understand why, see line (1) in Figure 4.162.

(This is the same line.) For any point Q on the loading curve, the line PQ has slope

QT

PT=

QT

PO + OT=

load

traveling time + searching time.

(b) The slope of the line PQ is maximized when the line is tangent to the loading curve, which happens with line (2).

The load is then approximately 7 worms.

(c) If the traveling time is increased, the point P moves to the left, to point P ′, say. If line (3) is tangent to the curve, it

will be tangent to the curve further to the right than line (2), so the optimal load is larger. This makes sense: if the

bird has to fly further, you’d expect it to bring back more worms each time.


P O

4

8 Number of worms

time

load(number of worms)

Figure 4.161

P ′ P

Traveling time

O T

Searching time

8

Q

Number of worms

(1)

(2)

(3)

time

load(number of worms)

Figure 4.162

56. (a) See Figure 4.163. The capacity appears to be 200 cars. The parking lot is full just before 8:30 am.

5:00 6:00 7:00 8:00 9:00

50

100

150

200

time (am)

total number of carsin parking lot

Figure 4.163

(b) The rate of arrival between 5:00 and 5:30 is (5 − 4)/0.5 = 2 cars/hours. We assign this the time 5:15 (rather than

either 5:00 or 5:30). The other values in Table 4.7 are calculated in a similar way. The data is plotted in Figure 4.164.

Table 4.7

Time (am) 5:15 5:45 6:15 6:45 7:15 7:45 8:15 8:45

Rate of arrival (cars/hour) 2 6 20 64 120 120 60 0

5:15 6:15 7:15 8:15

20

60

100

140

time (am)

rate of arrivalof cars

Figure 4.164

(c) Rush hour occurs around the time when the rate of arrival of cars is maximum, namely, about 7:30 am.

(d) The maximum of the rate of arrival occurs at the inflection point of the total number of cars in the lot.

57. For x > 0, the line in Figure 4.165 has

Slope =y

x=

x2e−3x

x= xe−3x.


If the slope has a maximum, it occurs where

d

dx(Slope) = 1 · e−3x − 3xe−3x = 0

e−3x (1 − 3x) = 0

x =1

3.

For this x-value,

Slope =1

3e−3(1/3) =

1

3e−1 =

1

3e.

Figure 4.165 shows that the slope tends toward 0 as x → ∞; the formula for the slope shows that the slope tends toward

0 as x → 0. Thus the only critical point, x = 1/3, must give a local and global maximum.

-� x

6?y

(x, x2e−3x)

x

y

Figure 4.165

58. (a) The rectangle in Figure 4.166 has area, A, given by

A = xy = xe−2x.

At a critical point, we have

dA

dx= 1 · e−2x − 2xe−2x = 0

e−2x (1 − 2x) = 0

x =1

2.

Since A = 0 when x = 0 and A → 0 as x → ∞, the critical point x = 1/2 is a local and global maximum. Thus

the maximum area is

A =1

2e−2(1/2) =

1

2e.

(b) The rectangle in Figure 4.166 has perimeter, P , given by

P = 2x + 2y = 2x + 2e−2x.

At a critical point, we have

dP

dx= 2 − 4e−2x = 0

e−2x =1

2

−2x = ln1

2

x = −1

2ln

1

2=

1

2ln 2.

To see if this critical point gives a maximum or minimum, we find

d2P

dx2= 8e−2x.

Since d2P/dx2 > 0 for all x, including x = 12

ln 2, the critical point is a local and global minimum. Thus, the

minimum perimeter is

P = 2(

1

2ln 2)

+ 2e−2( 12

ln 2) = ln 2 + 2e− ln 2 = ln 2 + 2 · 1

2= ln 2 + 1.


6

?

y

-� x

y = e−x/2

x

y

Figure 4.166

59. (a) To minimize A with h and s fixed we have to finddA

dθ.

dA

dθ=

3

2s2 d

dθ

(√3 − cos θ

sin θ

)

=3

2s2

(

sin2 θ − cos θ(√

3 − cos θ)

sin2 θ

)

=3

2s2

(

sin2 θ −√

3 cos θ + cos2 θ

sin2 θ

)

=3

2s2

(

1 −√

3 cos θ

sin2 θ

)

SetdA

dθ=

3

2s2

(

1 −√

3 cos θ

sin2 θ

)

= 0. Then 1 −√

3 cos θ = 0 and cos θ = 1√3

, so θ ≈ 54.7◦ is a critical

point of A. Since

d2A

dθ2

∣

∣

∣

∣

θ=54.7

=3

2s2

(√3 sin3 θ − 2 sin θ cos θ(1 −

√3 cos θ)

sin4 θ

)∣

∣

∣

∣

θ=54.7

≈ 3

2s2(2.122) > 0,

θ = 54.7 is indeed a minimum.

(b) Since 55◦ is very close to 54.7◦ , we conclude that bees attempt to minimize the surface areas of their honey combs.

60. (a) The vertical intercept is W = Ae−eb−c·0

= Ae−eb

. There is no horizontal intercept since the exponential function

is always positive. There is a horizontal asymptote. As t → ∞, we see that eb−ct = eb/ect → 0, since t is positive.

Therefore W → Ae0 = A, so there is a horizontal asymptote at W = A.

(b) The derivative isdW

dt= Ae−eb−ct

(−eb−ct)(−c) = Ace−eb−ct

eb−ct.

Thus, dW/dt is always positive, so W is always increasing and has no critical points. The second derivative is

d2W

dt2=

d

dt(Ace−eb−ct

)eb−ct + Ace−eb−ct d

dt(eb−ct)

= Ac2e−eb−ct

eb−cteb−ct + Ace−eb−ct

(−c)eb−ct

= Ac2e−eb−ct

eb−ct(eb−ct − 1).

Now eb−ct decreases from eb > 1 when t = 0 toward 0 as t → ∞. The second derivative changes sign from positive

to negative when eb−ct = 1, i.e., when b − ct = 0, or t = b/c. Thus the curve has an inflection point at t = b/c,

where W = Ae−eb−(b/c)c

= Ae−1.


CHECK YOUR UNDERSTANDING 281

A = 50, b = 2, c = 1

A = 50, b = 2, c = 5

A = 20, b = 2, c = 1

t

W

Figure 4.167

(d) The final size of the organism is given by the horizontal asymptote W = A. The curve is steepest at its inflection

point, which occurs at t = b/c, W = Ae−1. Since e = 2.71828 . . . ≈ 3, the size the organism when it is growing

fastest is about A/3, one third its final size. So yes, the Gompertz growth function is useful in modeling such growth.

CHECK YOUR UNDERSTANDING

1. False. The function f has a local minimum at p if f(p) ≤ f(x) for points x near p.

2. True—according to the definition, critical points of f occur where f ′(x) = 0 or f ′(x) is undefined.

3. False. The function f can also have critical points where f ′ is undefined.

4. True, since f has lower values to the left of p and to the right of p than it does at p.

5. False. If f ′(p) = 0 and f ′′(p) > 0 then f has a local minimum at p.

6. True. If f ′(p) = 0 and f ′′(p) > 0 then by the second derivative test f has a local minimum at p.

7. False. If f ′′(p) > 0, the function is concave up, but p must be a critical point before we can conclude that f has a local

minimum at p.

8. False. Consider f(x) = x3. The function has a critical point at x = 0, but x = 0 is neither a local maximum nor a local

minimum.

9. False. Consider f(x) = ex. The function and its derivative are defined for all x, and f ′(x) is never equal to zero. Thus, fhas no critical points.

10. False. Consider f(x) = |x|. The function has a local minimum at x = 0, but f ′(0) is undefined.

11. True, by the definition of inflection point.

12. False. Consider f(x) = x4. The second derivative is zero at x = 0, but the function is concave up for all x.

13. True. Consider f(x) = x3. The point x = 0 is both a critical point and an inflection point of f .

14. True. The graph of f(x) = x3 is concave down to the left of x = 0 and concave up to the right of x = 0, so there is an

inflection point at x = 0.

15. False. The function H(x) = x4 is concave up for all x, so H has no inflection points.

16. True. For example, the function f(x) = x3 + x has no critical points and has one inflection point, at x = 0.

17. True. For example, you can sketch the graph of such a function with a local maximum at (−1, 1), a local minimum at

(1,−1), and horizontal asymptote y = 0 as x → ±∞.

18. True. The second derivative is equal to zero at x = −1 and is positive to the left of x = −1 and negative to the right of

x = −1. Thus, f has an inflection point at x = −1.

19. False. The second derivative is equal to zero at x = −1 but the sign of f ′′ is negative both to the left and right of x = −1.

Thus, there is no change of concavity at x = −1.

20. True. Note that ex is positive for all x and that the factor (x + 1) changes sign from negative to positive at x = −1. Thus

the function changes concavity at x = −1 and x = −1 is an inflection point.

21. True. Consider f(x) = −x2. The function has a local and global maximum at x = 0.

22. False. The global maximum may also occur at an endpoint x = 1 or x = 2.

23. False. Consider f(x) = −x2 over the entire real line. There is no global minimum.

24. True. If f is increasing on the interval,then the largest value occurs at the right endpoint.


25. False. If f(x) < 0 then f is decreasing on the interval, and the largest value occurs at the left endpoint. The global

maximum will be at x = a.

26. True. For example, let S(x) = x. On the interval 1 ≤ x ≤ 2, the global maximum is 2 and occurs at the right-hand

endpoint x = 2. On the interval 2 ≤ x ≤ 3, the global maximum of 3 occurs at x = 3.

27. False. The function k has no critical points. As x → 0 from the right, the function increases without bound. Thus, k does

not have a global maximum when x > 0.

28. False. According to the definition, the function f has a global minimum at p on the interval given that f(p) ≤ f(x) for all

−5 ≤ x ≤ 5.

29. True, as defined in the text.

30. False. By the second derivative test, we have that p is a local minimum. However, a global minimum could occur at

another critical point or at an endpoint if the function is defined on a closed interval.

31. True. Cost is increasing at a slower rate than revenue–thus the cost of the next item will be less than the revenue for that

item.

32. True, since for the last item made and sold, we have received less in revenue than it cost to make, so by not making and

selling this item, profit would increase.

33. False, since a point where marginal revenue equals marginal cost may also indicate minimum profit.

34. True.

35. False. Maximum profit occurs at a critical point of the profit function—or at an endpoint of the profit function if there are

constraints on the number of items that can be made or sold.

36. True. When marginal profit is zero we have a critical point. The point must be tested to see if it gives a maximum of the

profit function.

37. False. The cost and revenue functions cross when a profit is turning into a loss, in which case profit is decreasing, or when

a loss is turning into a profit, in which case profit is increasing. In neither case is profit a maximum.

38. True. Since Profit = Revenue - Cost, when the cost and revenue curves are equal (and therefore the graphs cross), the

profit is zero.

39. True. Revenue is price times quantity. If the price is constant at price p, the graph of marginal revenue is the horizontal

line y = p.

40. False. If prices are constant at price p, the graph of revenue is a line with slope p.

41. True, as specified in the text.

42. True. The units of marginal cost and average cost have the same units—for example, if the units of cost are dollars, the

units of marginal cost and average cost are $/item.

43. False. Marginal cost gives the rate at which cost is changing—or, once q items have been produced, the approximate cost

of the next item. Average cost gives the average cost of producing q items. In general, these are not the same.

44. False. Average cost of q items can be visualized as the slope of a line from the origin to the point (q, C(q)) on the graph

of the cost function C.

45. False. If marginal cost is less than average cost, increasing production decreases average cost.

46. False. Marginal cost equals average cost at critical points of average cost.

47. True. If marginal cost is greater than average cost, increasing production increases average cost.

48. True, since the example in the section shows that the average cost function has critical points exactly where average cost

(which is the slope of the line from the origin to the cost function) is equal to the marginal cost (which is the slope of the

tangent to the cost function).

49. False. Not necessarily. Average cost may be increasing or decreasing as a functions of quantity, depending on the concavity

of the cost function.

50. False. Average cost is minimized when equal to marginal cost, but this is not the same quantity at which marginal cost is

minimized.

51. False. The elasticity of demand is given by E = |p/q · dq/dp| .52. False. If E > 1 then we say that demand is elastic.

53. False. If 0 ≤ E < 1 then we say that demand is inelastic.

54. True. This is what elasticity tells us.

PROJECTS FOR CHAPTER FOUR 283

55. False. If a product is considered a necessity, the demand is generally inelastic.

56. False. An increase in price may cause an increase or a decrease in revenue, depending on elasticity.

57. False. If elasticity E > 1, demand is elastic and revenue increases when price decreases.

58. True. At a critical point of the revenue function E = 1, and the demand is neither elastic nor inelastic.

59. True, since by increasing the price, demand drops, so costs are lower, but revenue is higher, so profit increases.

60. False. If E > 1 then increasing price causes a decrease in revenue, demand, and costs. We have seen examples where

profits decrease because revenue decreases more than costs.

61. True. The function P fits the model f(t) = L/(1 + Ce−kt) with L = 1000, C = 2, and k = 3.

62. False. Note that the end behavior of P does not approach the limiting value L. Instead, as t → ∞, the function values

approach zero. The fact that the exponent of e is not negative causes this behavior.

63. False. The function P has an inflection point when P = 1000/2 = 500. This occurs at approximately t = 0.2310.

64. True, since the logistic function P = L/(1 + Ce−kt) has an inflection point at L/2.

65. True. Note that as t increases, the denominator (1 + Ce−kt) approaches 1, so P (t) approaches L.

66. True, since the slope of the dose/response curve is the rate of change of the response as a function of the dose.

67. True. The logistic model approaches the carrying capacity as time increases.

68. True. If k is large the function approaches L more rapidly than when k is small.

69. True. At the inflection point, P = L/2, the function changes from concave up to concave down.

70. True. We can show this analytically by noting that P ′(t) = LkCe−kt(1 + Ce−kt)−2 > 0 for all t.

71. False. The formula for a surge function is y = ate−bt while the formula for an exponential decay function is y = ae−bt.

72. True.

73. False. The function is concave down and then concave up as it approaches the t-axis asymptotically.

74. True, the surge function will cross the line when it is increasing and again when it is decreasing.

75. False. The drug will never be effective since the concentration will always be below the minimum effective concentration.

76. True. There is exactly one inflection point, and it is always to the right of the one critical point.

77. True, as specified in the text.

78. True. Note that y can be written as y = (1/12)te−3t. This is a surge function with a = 1/12, b = 3.

79. False. The function y can be rewritten as y = (1/12)te3t. This function does not fit the model of a surge function because

the exponent of e is positive. Note that y → ∞ as t increases.

80. True, since the surge function y = ate−bt has maximum at t = 1/b.

PROJECTS FOR CHAPTER FOUR

1. Since a(q) = C(q)/q, we have C(q) = a(q) · q. Thus C′(q) = a′(q)q + a(q), and so

C′(q0) = a′(q0)q0 + a(q0).

Since t1 is the line tangent to a(q) at q = q0, the slope of t1 is a′(q0), and the equation of t1 is

y = a(q0) + a′(q0) · (q − q0) = a′(q0) · q + (a(q0) − a′(q0) · q0) .

Thus the y-intercept of t1 is given by a(q0) − a′(q0)q0, and the equation of the line t2 is

y = 2 · a′(q0) · q + (a(q0) − a′(q0) · q0)

since t2 has twice the slope of t1. Let’s compute the y-value on t2 when q = q0:

y = 2 · a′(q0) · q0 + (a(q0) − a′(q0) · q0) = a′(q0)q0 + a(q0) = C′(q0).

Hence C′(q0) is given by the point on t2 where q = q0.


2. (a) (i) We want to minimize A, the total area lost to the forest, which is made up of n firebreaks and 1 stand

of trees lying between firebreaks. The area of each firebreak is (50 km)(0.01 km) = 0.5 km2, so

the total area lost to the firebreaks is 0.5n km2. There are n total stands of trees between firebreaks.

The area of a single stand of trees can be found by subtracting the firebreak area from the forest and

dividing by n, so

Area of one stand of trees =2500− 0.5n

n.

Thus, the total area lost is

A = Area of one stand + Area lost to firebreaks

=2500 − 0.5n

n+ 0.5n =

2500

n− 0.5 + 0.5n.

We assume that A is a differentiable function of a continuous variable, n. Differentiating this function

yieldsdA

dn= −

2500

n2+ 0.5.

At critical points, dA/dn = 0, so 0.5 = 2500/n2 or n =√

2500/0.5 ≈ 70.7. Since n must be an

integer, we check that when n = 71, A = 70.211 and when n = 70, A = 70.214. Thus, n = 71 gives

a smaller area lost.

We can check that this is a local minimum since the second derivative is positive everywhere

d2A

dn2=

5000

n3> 0.

Finally, we check the endpoints: n = 1 yields the entire forest lost after a fire, since there is only

one stand of trees in this case and it all burns. The largest n is 5000, and in this case the firebreaks

remove the entire forest. Both of these cases maximize the area of forest lost. Thus, n = 71 is a global

minimum. So 71 firebreaks minimizes the area of forest lost.

(ii) Repeating the calculation using b for the width gives

A =2500

n− 50b + 50bn,

anddA

dn=

−2500

n2+ 50b,

with a critical point when b = 50/n2 so n =√

50/b. So, for example, if we make the width b four

times as large we need half as many firebreaks.

(b) We want to minimize A, the total area lost to the forest, which is made up of n firebreaks in one direction,

n firebreaks in the other, and one square of trees surrounded by firebreaks. The area of each firebreak is

0.5 km2, and there are 2n of them, giving a total of 0.5 · 2n. But this is larger than the total area covered

by the firebreaks, since it counts the small intersection squares, of size (0.01)2, twice. Since there are n2

intersections, we must subtract (0.01)2n2 from the total area of the 2n firebreaks. Thus,

Area covered by the firebreaks = 0.5 · 2n − (0.01)2n2.

To this we must add the area of one square patch of trees lost in a fire. These are squares of side (50 −

0.01n)/n = 50/n− 0.01. Thus the total area lost is

A = n − 0.0001n2 + (50/n− 0.01)2

Treating n as a continuous variable and differentiating this function yields

dA

dn= 1 − 0.0002n + 2

(

50

n− 0.01

)(

−50

n2

)

.

Using a computer algebra system to find critical points we find that dA/dn = 0 when n ≈ 17 and

n = 5000. Thus n = 17 gives a minimum lost area, since the endpoints of n = 1 and n = 5000 both yield

A = 2500 or the entire forest lost. So we use 17 firebreaks in each direction.

PROJECTS FOR CHAPTER FOUR 285

3. (a) Since the company can produce more goods if it has more raw materials to use, the function f(x) is

increasing. Thus, we expect the derivative f ′(x) to be positive.

(b) The cost to the company of acquiring x units of raw material is wx, and the revenue from the sale of f(x)units of the product is pf(x). The company’s profit is π(x) = Revenue − Cost = pf(x) − wx.

(c) Since profit π(x) is maximized at x = x∗, we have π′(x∗) = 0. From π′(x) = pf ′(x) − w, we have

pf ′(x∗) − w = 0. Thus f ′(x∗) = w/p.

(d) Computing the second derivative of π(x) gives π′′(x) = pf ′′(x). Since π(x) has a maximum at x = x∗,

the second derivative π′′(x∗) = pf ′′(x∗) is negative. Thus f ′′(x∗) is negative.

(e) Differentiate both sides of pf ′(x∗) − w = 0 with respect to w. The chain rule gives

pd

dwf ′(x∗) − 1 = 0

pf ′′(x∗)dx∗

dw− 1 = 0

dx∗

dw=

1

pf ′′(x∗).

Since f ′′(x∗) < 0, we see dx∗/dw is negative.

(f) Since dx∗/dw < 0, the quantity x∗ is a decreasing function of w. If the price w of the raw material goes

up, the company should buy less.

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